message stringlengths 2 48.6k | message_type stringclasses 2 values | message_id int64 0 1 | conversation_id int64 318 108k | cluster float64 8 8 | __index_level_0__ int64 636 217k |
|---|---|---|---|---|---|
Provide tags and a correct Python 3 solution for this coding contest problem.
There are famous Russian nesting dolls named matryoshkas sold in one of the souvenir stores nearby, and you'd like to buy several of them. The store has n different matryoshkas. Any matryoshka is a figure of volume out_i with an empty space inside of volume in_i (of course, out_i > in_i).
You don't have much free space inside your bag, but, fortunately, you know that matryoshkas can be nested one inside another. Formally, let's call a set of matryoshkas nested if we can rearrange dolls in such a way, that the first doll can be nested inside the second one, the second doll — inside the third one and so on. Matryoshka i can be nested inside matryoshka j if out_i ≤ in_j. So only the last doll will take space inside your bag.
Let's call extra space of a nested set of dolls as a total volume of empty space inside this structure. Obviously, it's equal to in_{i_1} + (in_{i_2} - out_{i_1}) + (in_{i_3} - out_{i_2}) + ... + (in_{i_k} - out_{i_{k-1}}), where i_1, i_2, ..., i_k are the indices of the chosen dolls in the order they are nested in each other.
Finally, let's call a nested subset of the given sequence as big enough if there isn't any doll from the sequence that can be added to the nested subset without breaking its nested property.
You want to buy many matryoshkas, so you should choose a big enough nested subset to buy it. But you will be disappointed if too much space in your bag will be wasted, so you want to choose a big enough subset so that its extra space is minimum possible among all big enough subsets. Now you wonder, how many different nested subsets meet these conditions (they are big enough, and there is no big enough subset such that its extra space is less than the extra space of the chosen subset). Two subsets are considered different if there exists at least one index i such that one of the subsets contains the i-th doll, and another subset doesn't.
Since the answer can be large, print it modulo 10^9 + 7.
Input
The first line contains a single integer n (1 ≤ n ≤ 2 ⋅ 10^5) — the number of matryoshkas.
The next n lines contain a description of each doll: two integers out_i and in_i (1 ≤ in_i < out_i ≤ 10^9) — the outer and inners volumes of the i-th matryoshka.
Output
Print one integer — the number of big enough nested subsets such that extra space of each of these subsets is minimum possible. Since the answer can be large, print it modulo 10^9 + 7.
Example
Input
7
4 1
4 2
4 2
2 1
5 4
6 4
3 2
Output
6
Note
There are 6 big enough nested subsets with minimum possible extra space in the example:
* \{1, 5\}: we can't add any other matryoshka and keep it nested; it's extra space is 1;
* \{1, 6\};
* \{2, 4, 5\};
* \{2, 4, 6\};
* \{3, 4, 5\};
* \{3, 4, 6\}.
There are no more "good" subsets because, for example, subset \{6, 7\} is not big enough (we can add the 4-th matryoshka to it) or subset \{4, 6, 7\} has extra space equal to 2. | instruction | 0 | 91,005 | 8 | 182,010 |
Tags: binary search, combinatorics, data structures, dp, shortest paths, sortings
Correct Solution:
```
import io, sys
input = lambda f=io.StringIO(sys.stdin.buffer.read().decode()).readline: f().rstrip()
from bisect import bisect_left as lb
MOD = 10 ** 9 + 7
n = int(input())
a = [tuple(map(int, input().split())) for _ in range(n)]
a = sorted((in_, out) for out, in_ in a)
dp = [None] * n
dp_min = [None] * n
for i in range(n - 1, -1, -1):
in_, out = a[i]
j = lb(a, (out, 0))
if j == n:
dp[i] = in_, 1
else:
empty, count = dp_min[j]
dp[i] = empty - out + in_, count
dp_min[i] = list(dp[i])
if i < n - 1:
if dp_min[i + 1][0] < dp_min[i][0]:
dp_min[i] = dp_min[i + 1]
elif dp_min[i + 1][0] == dp_min[i][0]:
dp_min[i][1] += dp_min[i + 1][1]
dp_min[i][1] %= MOD
print(dp_min[0][1])
``` | output | 1 | 91,005 | 8 | 182,011 |
Provide tags and a correct Python 3 solution for this coding contest problem.
There are famous Russian nesting dolls named matryoshkas sold in one of the souvenir stores nearby, and you'd like to buy several of them. The store has n different matryoshkas. Any matryoshka is a figure of volume out_i with an empty space inside of volume in_i (of course, out_i > in_i).
You don't have much free space inside your bag, but, fortunately, you know that matryoshkas can be nested one inside another. Formally, let's call a set of matryoshkas nested if we can rearrange dolls in such a way, that the first doll can be nested inside the second one, the second doll — inside the third one and so on. Matryoshka i can be nested inside matryoshka j if out_i ≤ in_j. So only the last doll will take space inside your bag.
Let's call extra space of a nested set of dolls as a total volume of empty space inside this structure. Obviously, it's equal to in_{i_1} + (in_{i_2} - out_{i_1}) + (in_{i_3} - out_{i_2}) + ... + (in_{i_k} - out_{i_{k-1}}), where i_1, i_2, ..., i_k are the indices of the chosen dolls in the order they are nested in each other.
Finally, let's call a nested subset of the given sequence as big enough if there isn't any doll from the sequence that can be added to the nested subset without breaking its nested property.
You want to buy many matryoshkas, so you should choose a big enough nested subset to buy it. But you will be disappointed if too much space in your bag will be wasted, so you want to choose a big enough subset so that its extra space is minimum possible among all big enough subsets. Now you wonder, how many different nested subsets meet these conditions (they are big enough, and there is no big enough subset such that its extra space is less than the extra space of the chosen subset). Two subsets are considered different if there exists at least one index i such that one of the subsets contains the i-th doll, and another subset doesn't.
Since the answer can be large, print it modulo 10^9 + 7.
Input
The first line contains a single integer n (1 ≤ n ≤ 2 ⋅ 10^5) — the number of matryoshkas.
The next n lines contain a description of each doll: two integers out_i and in_i (1 ≤ in_i < out_i ≤ 10^9) — the outer and inners volumes of the i-th matryoshka.
Output
Print one integer — the number of big enough nested subsets such that extra space of each of these subsets is minimum possible. Since the answer can be large, print it modulo 10^9 + 7.
Example
Input
7
4 1
4 2
4 2
2 1
5 4
6 4
3 2
Output
6
Note
There are 6 big enough nested subsets with minimum possible extra space in the example:
* \{1, 5\}: we can't add any other matryoshka and keep it nested; it's extra space is 1;
* \{1, 6\};
* \{2, 4, 5\};
* \{2, 4, 6\};
* \{3, 4, 5\};
* \{3, 4, 6\}.
There are no more "good" subsets because, for example, subset \{6, 7\} is not big enough (we can add the 4-th matryoshka to it) or subset \{4, 6, 7\} has extra space equal to 2. | instruction | 0 | 91,006 | 8 | 182,012 |
Tags: binary search, combinatorics, data structures, dp, shortest paths, sortings
Correct Solution:
```
import io, sys
input = lambda f=io.StringIO(sys.stdin.buffer.read().decode()).readline: f().rstrip()
from bisect import bisect_left as lb
MOD = 10 ** 9 + 7
n = int(input())
a = [tuple(map(int, input().split())) for _ in range(n)]
a = sorted((in_, out) for out, in_ in a)
dp_suf = [None] * n
for i in range(n - 1, -1, -1):
in_, out = a[i]
j = lb(a, (out, 0))
if j == n:
empty, count = in_, 1
else:
empty, count = dp_suf[j]
empty -= out - in_
if i < n - 1:
if empty > dp_suf[i + 1][0]:
empty, count = dp_suf[i + 1]
elif empty == dp_suf[i + 1][0]:
count += dp_suf[i + 1][1]
dp_suf[i] = empty, count % MOD
print(dp_suf[0][1])
``` | output | 1 | 91,006 | 8 | 182,013 |
Provide tags and a correct Python 3 solution for this coding contest problem.
There are famous Russian nesting dolls named matryoshkas sold in one of the souvenir stores nearby, and you'd like to buy several of them. The store has n different matryoshkas. Any matryoshka is a figure of volume out_i with an empty space inside of volume in_i (of course, out_i > in_i).
You don't have much free space inside your bag, but, fortunately, you know that matryoshkas can be nested one inside another. Formally, let's call a set of matryoshkas nested if we can rearrange dolls in such a way, that the first doll can be nested inside the second one, the second doll — inside the third one and so on. Matryoshka i can be nested inside matryoshka j if out_i ≤ in_j. So only the last doll will take space inside your bag.
Let's call extra space of a nested set of dolls as a total volume of empty space inside this structure. Obviously, it's equal to in_{i_1} + (in_{i_2} - out_{i_1}) + (in_{i_3} - out_{i_2}) + ... + (in_{i_k} - out_{i_{k-1}}), where i_1, i_2, ..., i_k are the indices of the chosen dolls in the order they are nested in each other.
Finally, let's call a nested subset of the given sequence as big enough if there isn't any doll from the sequence that can be added to the nested subset without breaking its nested property.
You want to buy many matryoshkas, so you should choose a big enough nested subset to buy it. But you will be disappointed if too much space in your bag will be wasted, so you want to choose a big enough subset so that its extra space is minimum possible among all big enough subsets. Now you wonder, how many different nested subsets meet these conditions (they are big enough, and there is no big enough subset such that its extra space is less than the extra space of the chosen subset). Two subsets are considered different if there exists at least one index i such that one of the subsets contains the i-th doll, and another subset doesn't.
Since the answer can be large, print it modulo 10^9 + 7.
Input
The first line contains a single integer n (1 ≤ n ≤ 2 ⋅ 10^5) — the number of matryoshkas.
The next n lines contain a description of each doll: two integers out_i and in_i (1 ≤ in_i < out_i ≤ 10^9) — the outer and inners volumes of the i-th matryoshka.
Output
Print one integer — the number of big enough nested subsets such that extra space of each of these subsets is minimum possible. Since the answer can be large, print it modulo 10^9 + 7.
Example
Input
7
4 1
4 2
4 2
2 1
5 4
6 4
3 2
Output
6
Note
There are 6 big enough nested subsets with minimum possible extra space in the example:
* \{1, 5\}: we can't add any other matryoshka and keep it nested; it's extra space is 1;
* \{1, 6\};
* \{2, 4, 5\};
* \{2, 4, 6\};
* \{3, 4, 5\};
* \{3, 4, 6\}.
There are no more "good" subsets because, for example, subset \{6, 7\} is not big enough (we can add the 4-th matryoshka to it) or subset \{4, 6, 7\} has extra space equal to 2. | instruction | 0 | 91,007 | 8 | 182,014 |
Tags: binary search, combinatorics, data structures, dp, shortest paths, sortings
Correct Solution:
```
mod = 1000000007
eps = 10**-9
def main():
import sys
input = sys.stdin.buffer.readline
N = int(input())
A = []
B = []
val = set()
val.add(0)
a_max = -1
for _ in range(N):
a, b = map(int, input().split())
a, b = b, a
A.append(a)
B.append(b)
val.add(a)
val.add(b)
a_max = max(a_max, a)
val = sorted(list(val))
val2idx = {v: i for i, v in enumerate(val)}
NN = len(val)
adj = [[] for _ in range(NN)]
for i in range(NN-1):
adj[i].append((i+1, val[i+1] - val[i]))
for a_, b_ in zip(A, B):
a = val2idx[a_]
b = val2idx[b_]
adj[a].append((b, 0))
dist = [10**10] * NN
dist[0] = 0
for i in range(NN-1):
for b, cost in adj[i]:
dist[b] = min(dist[b], dist[i] + cost)
min_space = 10**10
B_set = set(B)
for b in B_set:
if b <= a_max:
continue
ib = val2idx[b]
min_space = min(min_space, dist[ib])
dp = [0] * NN
dp[0] = 1
for i in range(NN - 1):
for b, cost in adj[i]:
if dist[i] + cost == dist[b]:
dp[b] = (dp[b] + dp[i])%mod
ans = 0
for b in B_set:
if b <= a_max:
continue
ib = val2idx[b]
if dist[ib] == min_space:
ans = (ans + dp[ib])%mod
print(ans)
if __name__ == '__main__':
main()
``` | output | 1 | 91,007 | 8 | 182,015 |
Provide tags and a correct Python 3 solution for this coding contest problem.
There are famous Russian nesting dolls named matryoshkas sold in one of the souvenir stores nearby, and you'd like to buy several of them. The store has n different matryoshkas. Any matryoshka is a figure of volume out_i with an empty space inside of volume in_i (of course, out_i > in_i).
You don't have much free space inside your bag, but, fortunately, you know that matryoshkas can be nested one inside another. Formally, let's call a set of matryoshkas nested if we can rearrange dolls in such a way, that the first doll can be nested inside the second one, the second doll — inside the third one and so on. Matryoshka i can be nested inside matryoshka j if out_i ≤ in_j. So only the last doll will take space inside your bag.
Let's call extra space of a nested set of dolls as a total volume of empty space inside this structure. Obviously, it's equal to in_{i_1} + (in_{i_2} - out_{i_1}) + (in_{i_3} - out_{i_2}) + ... + (in_{i_k} - out_{i_{k-1}}), where i_1, i_2, ..., i_k are the indices of the chosen dolls in the order they are nested in each other.
Finally, let's call a nested subset of the given sequence as big enough if there isn't any doll from the sequence that can be added to the nested subset without breaking its nested property.
You want to buy many matryoshkas, so you should choose a big enough nested subset to buy it. But you will be disappointed if too much space in your bag will be wasted, so you want to choose a big enough subset so that its extra space is minimum possible among all big enough subsets. Now you wonder, how many different nested subsets meet these conditions (they are big enough, and there is no big enough subset such that its extra space is less than the extra space of the chosen subset). Two subsets are considered different if there exists at least one index i such that one of the subsets contains the i-th doll, and another subset doesn't.
Since the answer can be large, print it modulo 10^9 + 7.
Input
The first line contains a single integer n (1 ≤ n ≤ 2 ⋅ 10^5) — the number of matryoshkas.
The next n lines contain a description of each doll: two integers out_i and in_i (1 ≤ in_i < out_i ≤ 10^9) — the outer and inners volumes of the i-th matryoshka.
Output
Print one integer — the number of big enough nested subsets such that extra space of each of these subsets is minimum possible. Since the answer can be large, print it modulo 10^9 + 7.
Example
Input
7
4 1
4 2
4 2
2 1
5 4
6 4
3 2
Output
6
Note
There are 6 big enough nested subsets with minimum possible extra space in the example:
* \{1, 5\}: we can't add any other matryoshka and keep it nested; it's extra space is 1;
* \{1, 6\};
* \{2, 4, 5\};
* \{2, 4, 6\};
* \{3, 4, 5\};
* \{3, 4, 6\}.
There are no more "good" subsets because, for example, subset \{6, 7\} is not big enough (we can add the 4-th matryoshka to it) or subset \{4, 6, 7\} has extra space equal to 2. | instruction | 0 | 91,008 | 8 | 182,016 |
Tags: binary search, combinatorics, data structures, dp, shortest paths, sortings
Correct Solution:
```
import sys
input = sys.stdin.readline
from bisect import bisect_right as br
P = 10**9+7
N = int(input())
X = []
maxinn = 0
for _ in range(N):
a, b = map(int, input().split())
maxinn = max(maxinn, b)
X.append((a, b))
X = sorted(X)
OUT = [0]
VOL = [0]
CNT = [1]
for out, inn in X:
i = br(OUT, inn) - 1
vol = VOL[i] + out - inn
if OUT[-1] != out:
OUT.append(out)
VOL.append(VOL[-1] if len(CNT)>1 else 0)
CNT.append(CNT[-1] if len(CNT)>1 else 0)
if VOL[-1] < vol:
VOL[-1] = vol
CNT[-1] = CNT[i]
elif VOL[-1] == vol:
CNT[-1] += CNT[i]
CNT[-1] %= P
mi = min([OUT[i]-VOL[i] for i in range(len(CNT)) if OUT[i] > maxinn])
print(sum([CNT[i] for i in range(len(CNT)) if OUT[i] > maxinn and OUT[i]-VOL[i] == mi])%P)
``` | output | 1 | 91,008 | 8 | 182,017 |
Provide tags and a correct Python 3 solution for this coding contest problem.
There are famous Russian nesting dolls named matryoshkas sold in one of the souvenir stores nearby, and you'd like to buy several of them. The store has n different matryoshkas. Any matryoshka is a figure of volume out_i with an empty space inside of volume in_i (of course, out_i > in_i).
You don't have much free space inside your bag, but, fortunately, you know that matryoshkas can be nested one inside another. Formally, let's call a set of matryoshkas nested if we can rearrange dolls in such a way, that the first doll can be nested inside the second one, the second doll — inside the third one and so on. Matryoshka i can be nested inside matryoshka j if out_i ≤ in_j. So only the last doll will take space inside your bag.
Let's call extra space of a nested set of dolls as a total volume of empty space inside this structure. Obviously, it's equal to in_{i_1} + (in_{i_2} - out_{i_1}) + (in_{i_3} - out_{i_2}) + ... + (in_{i_k} - out_{i_{k-1}}), where i_1, i_2, ..., i_k are the indices of the chosen dolls in the order they are nested in each other.
Finally, let's call a nested subset of the given sequence as big enough if there isn't any doll from the sequence that can be added to the nested subset without breaking its nested property.
You want to buy many matryoshkas, so you should choose a big enough nested subset to buy it. But you will be disappointed if too much space in your bag will be wasted, so you want to choose a big enough subset so that its extra space is minimum possible among all big enough subsets. Now you wonder, how many different nested subsets meet these conditions (they are big enough, and there is no big enough subset such that its extra space is less than the extra space of the chosen subset). Two subsets are considered different if there exists at least one index i such that one of the subsets contains the i-th doll, and another subset doesn't.
Since the answer can be large, print it modulo 10^9 + 7.
Input
The first line contains a single integer n (1 ≤ n ≤ 2 ⋅ 10^5) — the number of matryoshkas.
The next n lines contain a description of each doll: two integers out_i and in_i (1 ≤ in_i < out_i ≤ 10^9) — the outer and inners volumes of the i-th matryoshka.
Output
Print one integer — the number of big enough nested subsets such that extra space of each of these subsets is minimum possible. Since the answer can be large, print it modulo 10^9 + 7.
Example
Input
7
4 1
4 2
4 2
2 1
5 4
6 4
3 2
Output
6
Note
There are 6 big enough nested subsets with minimum possible extra space in the example:
* \{1, 5\}: we can't add any other matryoshka and keep it nested; it's extra space is 1;
* \{1, 6\};
* \{2, 4, 5\};
* \{2, 4, 6\};
* \{3, 4, 5\};
* \{3, 4, 6\}.
There are no more "good" subsets because, for example, subset \{6, 7\} is not big enough (we can add the 4-th matryoshka to it) or subset \{4, 6, 7\} has extra space equal to 2. | instruction | 0 | 91,009 | 8 | 182,018 |
Tags: binary search, combinatorics, data structures, dp, shortest paths, sortings
Correct Solution:
```
from bisect import bisect_left as lb
MOD = 10 ** 9 + 7
n = int(input())
a = [tuple(map(int, input().split())) for _ in range(n)]
a = sorted((in_, out) for out, in_ in a)
dp_suf = [None] * n
for i in range(n - 1, -1, -1):
in_, out = a[i]
j = lb(a, (out, 0))
if j == n:
empty, count = in_, 1
else:
empty, count = dp_suf[j]
empty -= out - in_
if i < n - 1:
if empty > dp_suf[i + 1][0]:
empty, count = dp_suf[i + 1]
elif empty == dp_suf[i + 1][0]:
count += dp_suf[i + 1][1]
dp_suf[i] = empty, count % MOD
print(dp_suf[0][1])
``` | output | 1 | 91,009 | 8 | 182,019 |
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response.
There are famous Russian nesting dolls named matryoshkas sold in one of the souvenir stores nearby, and you'd like to buy several of them. The store has n different matryoshkas. Any matryoshka is a figure of volume out_i with an empty space inside of volume in_i (of course, out_i > in_i).
You don't have much free space inside your bag, but, fortunately, you know that matryoshkas can be nested one inside another. Formally, let's call a set of matryoshkas nested if we can rearrange dolls in such a way, that the first doll can be nested inside the second one, the second doll — inside the third one and so on. Matryoshka i can be nested inside matryoshka j if out_i ≤ in_j. So only the last doll will take space inside your bag.
Let's call extra space of a nested set of dolls as a total volume of empty space inside this structure. Obviously, it's equal to in_{i_1} + (in_{i_2} - out_{i_1}) + (in_{i_3} - out_{i_2}) + ... + (in_{i_k} - out_{i_{k-1}}), where i_1, i_2, ..., i_k are the indices of the chosen dolls in the order they are nested in each other.
Finally, let's call a nested subset of the given sequence as big enough if there isn't any doll from the sequence that can be added to the nested subset without breaking its nested property.
You want to buy many matryoshkas, so you should choose a big enough nested subset to buy it. But you will be disappointed if too much space in your bag will be wasted, so you want to choose a big enough subset so that its extra space is minimum possible among all big enough subsets. Now you wonder, how many different nested subsets meet these conditions (they are big enough, and there is no big enough subset such that its extra space is less than the extra space of the chosen subset). Two subsets are considered different if there exists at least one index i such that one of the subsets contains the i-th doll, and another subset doesn't.
Since the answer can be large, print it modulo 10^9 + 7.
Input
The first line contains a single integer n (1 ≤ n ≤ 2 ⋅ 10^5) — the number of matryoshkas.
The next n lines contain a description of each doll: two integers out_i and in_i (1 ≤ in_i < out_i ≤ 10^9) — the outer and inners volumes of the i-th matryoshka.
Output
Print one integer — the number of big enough nested subsets such that extra space of each of these subsets is minimum possible. Since the answer can be large, print it modulo 10^9 + 7.
Example
Input
7
4 1
4 2
4 2
2 1
5 4
6 4
3 2
Output
6
Note
There are 6 big enough nested subsets with minimum possible extra space in the example:
* \{1, 5\}: we can't add any other matryoshka and keep it nested; it's extra space is 1;
* \{1, 6\};
* \{2, 4, 5\};
* \{2, 4, 6\};
* \{3, 4, 5\};
* \{3, 4, 6\}.
There are no more "good" subsets because, for example, subset \{6, 7\} is not big enough (we can add the 4-th matryoshka to it) or subset \{4, 6, 7\} has extra space equal to 2.
Submitted Solution:
```
from bisect import bisect_right, bisect_left
n = int(input())
gout = []
alls = []
mout = 0
for i in range(n):
a, b = map(int, input().split())
gout.append((a, b, i))
alls.append((a, b, i))
mout = max(mout, b)
gout.sort()
presets = [[]]
sus = []
for i in range(n):
sus.append(gout[i])
presets.append(sus.copy())
graph = {}
for a, b, i in alls:
z = bisect_right(gout, (b, 10 ** 10, 10 ** 10))
graph[(a, b, i)] = presets[z]
ans = 0
for y in graph:
if y[0] > mout:
h = graph[y]
b = y[1]
z = bisect_right(h, (b, 10 ** 10, 10 ** 10))
jz = bisect_left(h, (b, -10 ** 10, -10 ** 10))
ans += z - jz
ans %= (10 ** 9 + 7)
print(ans % (10 ** 9 + 7))
``` | instruction | 0 | 91,010 | 8 | 182,020 |
No | output | 1 | 91,010 | 8 | 182,021 |
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response.
There are famous Russian nesting dolls named matryoshkas sold in one of the souvenir stores nearby, and you'd like to buy several of them. The store has n different matryoshkas. Any matryoshka is a figure of volume out_i with an empty space inside of volume in_i (of course, out_i > in_i).
You don't have much free space inside your bag, but, fortunately, you know that matryoshkas can be nested one inside another. Formally, let's call a set of matryoshkas nested if we can rearrange dolls in such a way, that the first doll can be nested inside the second one, the second doll — inside the third one and so on. Matryoshka i can be nested inside matryoshka j if out_i ≤ in_j. So only the last doll will take space inside your bag.
Let's call extra space of a nested set of dolls as a total volume of empty space inside this structure. Obviously, it's equal to in_{i_1} + (in_{i_2} - out_{i_1}) + (in_{i_3} - out_{i_2}) + ... + (in_{i_k} - out_{i_{k-1}}), where i_1, i_2, ..., i_k are the indices of the chosen dolls in the order they are nested in each other.
Finally, let's call a nested subset of the given sequence as big enough if there isn't any doll from the sequence that can be added to the nested subset without breaking its nested property.
You want to buy many matryoshkas, so you should choose a big enough nested subset to buy it. But you will be disappointed if too much space in your bag will be wasted, so you want to choose a big enough subset so that its extra space is minimum possible among all big enough subsets. Now you wonder, how many different nested subsets meet these conditions (they are big enough, and there is no big enough subset such that its extra space is less than the extra space of the chosen subset). Two subsets are considered different if there exists at least one index i such that one of the subsets contains the i-th doll, and another subset doesn't.
Since the answer can be large, print it modulo 10^9 + 7.
Input
The first line contains a single integer n (1 ≤ n ≤ 2 ⋅ 10^5) — the number of matryoshkas.
The next n lines contain a description of each doll: two integers out_i and in_i (1 ≤ in_i < out_i ≤ 10^9) — the outer and inners volumes of the i-th matryoshka.
Output
Print one integer — the number of big enough nested subsets such that extra space of each of these subsets is minimum possible. Since the answer can be large, print it modulo 10^9 + 7.
Example
Input
7
4 1
4 2
4 2
2 1
5 4
6 4
3 2
Output
6
Note
There are 6 big enough nested subsets with minimum possible extra space in the example:
* \{1, 5\}: we can't add any other matryoshka and keep it nested; it's extra space is 1;
* \{1, 6\};
* \{2, 4, 5\};
* \{2, 4, 6\};
* \{3, 4, 5\};
* \{3, 4, 6\}.
There are no more "good" subsets because, for example, subset \{6, 7\} is not big enough (we can add the 4-th matryoshka to it) or subset \{4, 6, 7\} has extra space equal to 2.
Submitted Solution:
```
MOD = 10**9 + 7
n = int(input())
mat = []
vset = set()
for i in range(n):
out, inn = [int(item) for item in input().split()]
mat.append((inn, out))
vset.add(inn)
vset.add(out)
mat.sort()
vset = list(vset)
vset.sort()
vdic = dict()
for i, item in enumerate(vset):
vdic[item] = i
# print(mat)
# print(vdic)
edge = [[] for _ in range(len(vdic))]
for inn, out in mat:
edge[vdic[inn]].append(vdic[out])
dp = [0] * len(vdic)
survive = [0] * len(vdic)
dp[0] = 1
ans = 0
max_to = 0
for i, l in enumerate(edge):
for j in l:
dp[j] += dp[i]
dp[j] %= MOD
if dp[i] != 0:
max_to = max(max_to, j)
if i < len(edge)-1 and len(l) == 0 and max_to == i:
dp[i+1] = dp[i]
survive[i+1] = 1
print(dp)
for i, l in enumerate(edge[::-1]):
if len(l) == 0:
ans += dp[len(vdic) - 1 - i]
ans %= MOD
if survive[len(vdic) - 1 - i] == 1:
break
else:
if dp[len(vdic) - 1 - i] == 0:
continue
else:
break
print(ans)
``` | instruction | 0 | 91,011 | 8 | 182,022 |
No | output | 1 | 91,011 | 8 | 182,023 |
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response.
There are famous Russian nesting dolls named matryoshkas sold in one of the souvenir stores nearby, and you'd like to buy several of them. The store has n different matryoshkas. Any matryoshka is a figure of volume out_i with an empty space inside of volume in_i (of course, out_i > in_i).
You don't have much free space inside your bag, but, fortunately, you know that matryoshkas can be nested one inside another. Formally, let's call a set of matryoshkas nested if we can rearrange dolls in such a way, that the first doll can be nested inside the second one, the second doll — inside the third one and so on. Matryoshka i can be nested inside matryoshka j if out_i ≤ in_j. So only the last doll will take space inside your bag.
Let's call extra space of a nested set of dolls as a total volume of empty space inside this structure. Obviously, it's equal to in_{i_1} + (in_{i_2} - out_{i_1}) + (in_{i_3} - out_{i_2}) + ... + (in_{i_k} - out_{i_{k-1}}), where i_1, i_2, ..., i_k are the indices of the chosen dolls in the order they are nested in each other.
Finally, let's call a nested subset of the given sequence as big enough if there isn't any doll from the sequence that can be added to the nested subset without breaking its nested property.
You want to buy many matryoshkas, so you should choose a big enough nested subset to buy it. But you will be disappointed if too much space in your bag will be wasted, so you want to choose a big enough subset so that its extra space is minimum possible among all big enough subsets. Now you wonder, how many different nested subsets meet these conditions (they are big enough, and there is no big enough subset such that its extra space is less than the extra space of the chosen subset). Two subsets are considered different if there exists at least one index i such that one of the subsets contains the i-th doll, and another subset doesn't.
Since the answer can be large, print it modulo 10^9 + 7.
Input
The first line contains a single integer n (1 ≤ n ≤ 2 ⋅ 10^5) — the number of matryoshkas.
The next n lines contain a description of each doll: two integers out_i and in_i (1 ≤ in_i < out_i ≤ 10^9) — the outer and inners volumes of the i-th matryoshka.
Output
Print one integer — the number of big enough nested subsets such that extra space of each of these subsets is minimum possible. Since the answer can be large, print it modulo 10^9 + 7.
Example
Input
7
4 1
4 2
4 2
2 1
5 4
6 4
3 2
Output
6
Note
There are 6 big enough nested subsets with minimum possible extra space in the example:
* \{1, 5\}: we can't add any other matryoshka and keep it nested; it's extra space is 1;
* \{1, 6\};
* \{2, 4, 5\};
* \{2, 4, 6\};
* \{3, 4, 5\};
* \{3, 4, 6\}.
There are no more "good" subsets because, for example, subset \{6, 7\} is not big enough (we can add the 4-th matryoshka to it) or subset \{4, 6, 7\} has extra space equal to 2.
Submitted Solution:
```
MOD = 10**9 + 7
n = int(input())
mat = []
vset = set()
for i in range(n):
out, inn = [int(item) for item in input().split()]
mat.append((inn, out))
vset.add(inn)
vset.add(out)
mat.sort()
vset = list(vset)
vset.sort()
vdic = dict()
for i, item in enumerate(vset):
vdic[item] = i
# print(mat)
# print(vdic)
edge = [[] for _ in range(len(vdic))]
for inn, out in mat:
edge[vdic[inn]].append(vdic[out])
# print(edge)
dp = [0] * len(vdic)
dp[0] = 1
ans = 0
for i, l in enumerate(edge):
for j in l:
dp[j] += dp[i]
dp[j] %= MOD
for i, l in enumerate(edge[::-1]):
if len(l) == 0:
ans += dp[len(vdic) - 1 - i]
ans %= MOD
else:
break
print(ans)
``` | instruction | 0 | 91,012 | 8 | 182,024 |
No | output | 1 | 91,012 | 8 | 182,025 |
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response.
There are famous Russian nesting dolls named matryoshkas sold in one of the souvenir stores nearby, and you'd like to buy several of them. The store has n different matryoshkas. Any matryoshka is a figure of volume out_i with an empty space inside of volume in_i (of course, out_i > in_i).
You don't have much free space inside your bag, but, fortunately, you know that matryoshkas can be nested one inside another. Formally, let's call a set of matryoshkas nested if we can rearrange dolls in such a way, that the first doll can be nested inside the second one, the second doll — inside the third one and so on. Matryoshka i can be nested inside matryoshka j if out_i ≤ in_j. So only the last doll will take space inside your bag.
Let's call extra space of a nested set of dolls as a total volume of empty space inside this structure. Obviously, it's equal to in_{i_1} + (in_{i_2} - out_{i_1}) + (in_{i_3} - out_{i_2}) + ... + (in_{i_k} - out_{i_{k-1}}), where i_1, i_2, ..., i_k are the indices of the chosen dolls in the order they are nested in each other.
Finally, let's call a nested subset of the given sequence as big enough if there isn't any doll from the sequence that can be added to the nested subset without breaking its nested property.
You want to buy many matryoshkas, so you should choose a big enough nested subset to buy it. But you will be disappointed if too much space in your bag will be wasted, so you want to choose a big enough subset so that its extra space is minimum possible among all big enough subsets. Now you wonder, how many different nested subsets meet these conditions (they are big enough, and there is no big enough subset such that its extra space is less than the extra space of the chosen subset). Two subsets are considered different if there exists at least one index i such that one of the subsets contains the i-th doll, and another subset doesn't.
Since the answer can be large, print it modulo 10^9 + 7.
Input
The first line contains a single integer n (1 ≤ n ≤ 2 ⋅ 10^5) — the number of matryoshkas.
The next n lines contain a description of each doll: two integers out_i and in_i (1 ≤ in_i < out_i ≤ 10^9) — the outer and inners volumes of the i-th matryoshka.
Output
Print one integer — the number of big enough nested subsets such that extra space of each of these subsets is minimum possible. Since the answer can be large, print it modulo 10^9 + 7.
Example
Input
7
4 1
4 2
4 2
2 1
5 4
6 4
3 2
Output
6
Note
There are 6 big enough nested subsets with minimum possible extra space in the example:
* \{1, 5\}: we can't add any other matryoshka and keep it nested; it's extra space is 1;
* \{1, 6\};
* \{2, 4, 5\};
* \{2, 4, 6\};
* \{3, 4, 5\};
* \{3, 4, 6\}.
There are no more "good" subsets because, for example, subset \{6, 7\} is not big enough (we can add the 4-th matryoshka to it) or subset \{4, 6, 7\} has extra space equal to 2.
Submitted Solution:
```
from bisect import bisect_right, bisect_left
n = int(input())
gout = []
alls = []
mout = 0
A = []
B = []
for i in range(n):
a, b = map(int, input().split())
gout.append((a, b, i))
alls.append((a, b, i))
mout = max(mout, b)
A.append(a)
B.append(b)
A.sort()
B.sort()
kr = 10 ** 9
u1 = 0
u2 = 0
while u1 < len(A) and u2 < len(A):
if A[u1] <= B[u2]:
kr = min(kr, B[u2] - A[u1])
u1 += 1
else:
u2 += 1
if kr == 10 ** 9:
print(0)
else:
gout.sort()
presets = [[]]
sus = []
for i in range(n):
sus.append(gout[i])
presets.append(sus.copy())
graph = {}
for a, b, i in alls:
z = bisect_right(gout, (b, 10 ** 10, 10 ** 10))
graph[(a, b, i)] = presets[z]
ans = 0
for y in graph:
if y[0] > mout:
h = graph[y]
b = y[1]
z = bisect_right(h, (b - kr, 10 ** 10, 10 ** 10))
jz = bisect_left(h, (b, -10 ** 10, -10 ** 10))
ans += z - jz
ans %= (10 ** 9 + 7)
print(ans % (10 ** 9 + 7))
``` | instruction | 0 | 91,013 | 8 | 182,026 |
No | output | 1 | 91,013 | 8 | 182,027 |
Provide tags and a correct Python 3 solution for this coding contest problem.
Fox Ciel has n boxes in her room. They have the same size and weight, but they might have different strength. The i-th box can hold at most xi boxes on its top (we'll call xi the strength of the box).
Since all the boxes have the same size, Ciel cannot put more than one box directly on the top of some box. For example, imagine Ciel has three boxes: the first has strength 2, the second has strength 1 and the third has strength 1. She cannot put the second and the third box simultaneously directly on the top of the first one. But she can put the second box directly on the top of the first one, and then the third box directly on the top of the second one. We will call such a construction of boxes a pile.
<image>
Fox Ciel wants to construct piles from all the boxes. Each pile will contain some boxes from top to bottom, and there cannot be more than xi boxes on the top of i-th box. What is the minimal number of piles she needs to construct?
Input
The first line contains an integer n (1 ≤ n ≤ 100). The next line contains n integers x1, x2, ..., xn (0 ≤ xi ≤ 100).
Output
Output a single integer — the minimal possible number of piles.
Examples
Input
3
0 0 10
Output
2
Input
5
0 1 2 3 4
Output
1
Input
4
0 0 0 0
Output
4
Input
9
0 1 0 2 0 1 1 2 10
Output
3
Note
In example 1, one optimal way is to build 2 piles: the first pile contains boxes 1 and 3 (from top to bottom), the second pile contains only box 2.
<image>
In example 2, we can build only 1 pile that contains boxes 1, 2, 3, 4, 5 (from top to bottom).
<image> | instruction | 0 | 91,241 | 8 | 182,482 |
Tags: greedy, sortings
Correct Solution:
```
def validate_stack(stack):
for i in range(len(stack)):
if stack[i] < len(stack)-i-1:
return False
return True
# print(validate_stack([4,4,4,4,4]))
if __name__ == '__main__':
n = int(input())
box_strength = [int(x) for x in input().split()]
box_strength = sorted(box_strength, reverse = True)
flag = True
ans = 100
for i in range(1,n+1):
matrix = [[] for x in range(i)]
flag = True
ans = i
# print(matrix)
for idx, j in enumerate(box_strength):
# print(idx%i,j)
matrix[idx%i].append(j)
# print(matrix)
for stack in matrix:
if not validate_stack(stack):
flag = False
break
if flag:
break
print(ans)
``` | output | 1 | 91,241 | 8 | 182,483 |
Provide tags and a correct Python 3 solution for this coding contest problem.
Fox Ciel has n boxes in her room. They have the same size and weight, but they might have different strength. The i-th box can hold at most xi boxes on its top (we'll call xi the strength of the box).
Since all the boxes have the same size, Ciel cannot put more than one box directly on the top of some box. For example, imagine Ciel has three boxes: the first has strength 2, the second has strength 1 and the third has strength 1. She cannot put the second and the third box simultaneously directly on the top of the first one. But she can put the second box directly on the top of the first one, and then the third box directly on the top of the second one. We will call such a construction of boxes a pile.
<image>
Fox Ciel wants to construct piles from all the boxes. Each pile will contain some boxes from top to bottom, and there cannot be more than xi boxes on the top of i-th box. What is the minimal number of piles she needs to construct?
Input
The first line contains an integer n (1 ≤ n ≤ 100). The next line contains n integers x1, x2, ..., xn (0 ≤ xi ≤ 100).
Output
Output a single integer — the minimal possible number of piles.
Examples
Input
3
0 0 10
Output
2
Input
5
0 1 2 3 4
Output
1
Input
4
0 0 0 0
Output
4
Input
9
0 1 0 2 0 1 1 2 10
Output
3
Note
In example 1, one optimal way is to build 2 piles: the first pile contains boxes 1 and 3 (from top to bottom), the second pile contains only box 2.
<image>
In example 2, we can build only 1 pile that contains boxes 1, 2, 3, 4, 5 (from top to bottom).
<image> | instruction | 0 | 91,242 | 8 | 182,484 |
Tags: greedy, sortings
Correct Solution:
```
def go():
n = int(input())
a = [int(i) for i in input().split(' ')]
a.sort()
o = 1
for i in range(n):
if(a[i] < i // o):
o += 1
return o
print(go())
``` | output | 1 | 91,242 | 8 | 182,485 |
Provide tags and a correct Python 3 solution for this coding contest problem.
Fox Ciel has n boxes in her room. They have the same size and weight, but they might have different strength. The i-th box can hold at most xi boxes on its top (we'll call xi the strength of the box).
Since all the boxes have the same size, Ciel cannot put more than one box directly on the top of some box. For example, imagine Ciel has three boxes: the first has strength 2, the second has strength 1 and the third has strength 1. She cannot put the second and the third box simultaneously directly on the top of the first one. But she can put the second box directly on the top of the first one, and then the third box directly on the top of the second one. We will call such a construction of boxes a pile.
<image>
Fox Ciel wants to construct piles from all the boxes. Each pile will contain some boxes from top to bottom, and there cannot be more than xi boxes on the top of i-th box. What is the minimal number of piles she needs to construct?
Input
The first line contains an integer n (1 ≤ n ≤ 100). The next line contains n integers x1, x2, ..., xn (0 ≤ xi ≤ 100).
Output
Output a single integer — the minimal possible number of piles.
Examples
Input
3
0 0 10
Output
2
Input
5
0 1 2 3 4
Output
1
Input
4
0 0 0 0
Output
4
Input
9
0 1 0 2 0 1 1 2 10
Output
3
Note
In example 1, one optimal way is to build 2 piles: the first pile contains boxes 1 and 3 (from top to bottom), the second pile contains only box 2.
<image>
In example 2, we can build only 1 pile that contains boxes 1, 2, 3, 4, 5 (from top to bottom).
<image> | instruction | 0 | 91,243 | 8 | 182,486 |
Tags: greedy, sortings
Correct Solution:
```
import bisect
n = int(input())
xi = list(sorted(map(int, input().split())))
s = set(xi)
li = []
while len(xi) > 0:
li.append([xi.pop(0)])
i = 0
while i < len(xi):
if xi[i] >= len(li[-1]):
li[-1].append(xi.pop(i))
else:
i += 1
# print(li)
print(len(li))
``` | output | 1 | 91,243 | 8 | 182,487 |
Provide tags and a correct Python 3 solution for this coding contest problem.
Fox Ciel has n boxes in her room. They have the same size and weight, but they might have different strength. The i-th box can hold at most xi boxes on its top (we'll call xi the strength of the box).
Since all the boxes have the same size, Ciel cannot put more than one box directly on the top of some box. For example, imagine Ciel has three boxes: the first has strength 2, the second has strength 1 and the third has strength 1. She cannot put the second and the third box simultaneously directly on the top of the first one. But she can put the second box directly on the top of the first one, and then the third box directly on the top of the second one. We will call such a construction of boxes a pile.
<image>
Fox Ciel wants to construct piles from all the boxes. Each pile will contain some boxes from top to bottom, and there cannot be more than xi boxes on the top of i-th box. What is the minimal number of piles she needs to construct?
Input
The first line contains an integer n (1 ≤ n ≤ 100). The next line contains n integers x1, x2, ..., xn (0 ≤ xi ≤ 100).
Output
Output a single integer — the minimal possible number of piles.
Examples
Input
3
0 0 10
Output
2
Input
5
0 1 2 3 4
Output
1
Input
4
0 0 0 0
Output
4
Input
9
0 1 0 2 0 1 1 2 10
Output
3
Note
In example 1, one optimal way is to build 2 piles: the first pile contains boxes 1 and 3 (from top to bottom), the second pile contains only box 2.
<image>
In example 2, we can build only 1 pile that contains boxes 1, 2, 3, 4, 5 (from top to bottom).
<image> | instruction | 0 | 91,244 | 8 | 182,488 |
Tags: greedy, sortings
Correct Solution:
```
import math
class CodeforcesTask388ASolution:
def __init__(self):
self.result = ''
self.n = 0
self.boxes = []
def read_input(self):
self.n = int(input())
self.boxes = [int(x) for x in input().split(" ")]
def process_task(self):
counts = [self.boxes.count(x) for x in range(max(self.boxes) + 1)]
constraints = [math.ceil(sum(counts[0:x + 1]) / (x + 1)) for x in range(len(counts))]
self.result = str(max(constraints))
def get_result(self):
return self.result
if __name__ == "__main__":
Solution = CodeforcesTask388ASolution()
Solution.read_input()
Solution.process_task()
print(Solution.get_result())
``` | output | 1 | 91,244 | 8 | 182,489 |
Provide tags and a correct Python 3 solution for this coding contest problem.
Fox Ciel has n boxes in her room. They have the same size and weight, but they might have different strength. The i-th box can hold at most xi boxes on its top (we'll call xi the strength of the box).
Since all the boxes have the same size, Ciel cannot put more than one box directly on the top of some box. For example, imagine Ciel has three boxes: the first has strength 2, the second has strength 1 and the third has strength 1. She cannot put the second and the third box simultaneously directly on the top of the first one. But she can put the second box directly on the top of the first one, and then the third box directly on the top of the second one. We will call such a construction of boxes a pile.
<image>
Fox Ciel wants to construct piles from all the boxes. Each pile will contain some boxes from top to bottom, and there cannot be more than xi boxes on the top of i-th box. What is the minimal number of piles she needs to construct?
Input
The first line contains an integer n (1 ≤ n ≤ 100). The next line contains n integers x1, x2, ..., xn (0 ≤ xi ≤ 100).
Output
Output a single integer — the minimal possible number of piles.
Examples
Input
3
0 0 10
Output
2
Input
5
0 1 2 3 4
Output
1
Input
4
0 0 0 0
Output
4
Input
9
0 1 0 2 0 1 1 2 10
Output
3
Note
In example 1, one optimal way is to build 2 piles: the first pile contains boxes 1 and 3 (from top to bottom), the second pile contains only box 2.
<image>
In example 2, we can build only 1 pile that contains boxes 1, 2, 3, 4, 5 (from top to bottom).
<image> | instruction | 0 | 91,245 | 8 | 182,490 |
Tags: greedy, sortings
Correct Solution:
```
n=int(input())
x=list(map(int,input().split()))
x.sort()
nu=0
ans=0
mark=[]
for i in range(0,n):
mark.append(0)
for i in range(0,n):
fail=1
nu=0
for j in range(0,n):
if mark[j] == 0:
fail = 0
if x[j] >= nu:
nu+=1
mark[j]=1
if fail == 0:
ans+=1
else:
break
print(ans)
``` | output | 1 | 91,245 | 8 | 182,491 |
Provide tags and a correct Python 3 solution for this coding contest problem.
Fox Ciel has n boxes in her room. They have the same size and weight, but they might have different strength. The i-th box can hold at most xi boxes on its top (we'll call xi the strength of the box).
Since all the boxes have the same size, Ciel cannot put more than one box directly on the top of some box. For example, imagine Ciel has three boxes: the first has strength 2, the second has strength 1 and the third has strength 1. She cannot put the second and the third box simultaneously directly on the top of the first one. But she can put the second box directly on the top of the first one, and then the third box directly on the top of the second one. We will call such a construction of boxes a pile.
<image>
Fox Ciel wants to construct piles from all the boxes. Each pile will contain some boxes from top to bottom, and there cannot be more than xi boxes on the top of i-th box. What is the minimal number of piles she needs to construct?
Input
The first line contains an integer n (1 ≤ n ≤ 100). The next line contains n integers x1, x2, ..., xn (0 ≤ xi ≤ 100).
Output
Output a single integer — the minimal possible number of piles.
Examples
Input
3
0 0 10
Output
2
Input
5
0 1 2 3 4
Output
1
Input
4
0 0 0 0
Output
4
Input
9
0 1 0 2 0 1 1 2 10
Output
3
Note
In example 1, one optimal way is to build 2 piles: the first pile contains boxes 1 and 3 (from top to bottom), the second pile contains only box 2.
<image>
In example 2, we can build only 1 pile that contains boxes 1, 2, 3, 4, 5 (from top to bottom).
<image> | instruction | 0 | 91,246 | 8 | 182,492 |
Tags: greedy, sortings
Correct Solution:
```
n = int(input())
a = [int(i) for i in input().split(' ')]
a.sort()
s = 1
for i in range(n):
if(a[i] < i // s):
s += 1
print(s)
``` | output | 1 | 91,246 | 8 | 182,493 |
Provide tags and a correct Python 3 solution for this coding contest problem.
Fox Ciel has n boxes in her room. They have the same size and weight, but they might have different strength. The i-th box can hold at most xi boxes on its top (we'll call xi the strength of the box).
Since all the boxes have the same size, Ciel cannot put more than one box directly on the top of some box. For example, imagine Ciel has three boxes: the first has strength 2, the second has strength 1 and the third has strength 1. She cannot put the second and the third box simultaneously directly on the top of the first one. But she can put the second box directly on the top of the first one, and then the third box directly on the top of the second one. We will call such a construction of boxes a pile.
<image>
Fox Ciel wants to construct piles from all the boxes. Each pile will contain some boxes from top to bottom, and there cannot be more than xi boxes on the top of i-th box. What is the minimal number of piles she needs to construct?
Input
The first line contains an integer n (1 ≤ n ≤ 100). The next line contains n integers x1, x2, ..., xn (0 ≤ xi ≤ 100).
Output
Output a single integer — the minimal possible number of piles.
Examples
Input
3
0 0 10
Output
2
Input
5
0 1 2 3 4
Output
1
Input
4
0 0 0 0
Output
4
Input
9
0 1 0 2 0 1 1 2 10
Output
3
Note
In example 1, one optimal way is to build 2 piles: the first pile contains boxes 1 and 3 (from top to bottom), the second pile contains only box 2.
<image>
In example 2, we can build only 1 pile that contains boxes 1, 2, 3, 4, 5 (from top to bottom).
<image> | instruction | 0 | 91,247 | 8 | 182,494 |
Tags: greedy, sortings
Correct Solution:
```
n=int(input())
l=list(map(int,input().split()))
l=sorted(l)
l=l[::-1]
l1=[0]*n
k=0
for i in range(n) :
if l1[i]!=1 :
t=l[i]
p=1
r=0
l1[i]==1
V=[t]
for j in range(n) :
if l1[j]==0 and l[j]<t :
t=l[j]
l1[j]=1
V.append(t)
r=r+1
for j in range(n-1,-1,-1) :
if l1[j]!=1 :
if l[j] in V :
q=V.index(l[j])+1
if len(V)-q+1<=l[j] :
c=0
s=-1
u=len(V)-q+1
for e in range(q-1,-1,-1) :
if V[e]>=u+s :
s=s+1
else :
c=1
break
if c==0 :
l1[j]=1
V=V[:q-1]+[l[j]]+V[q-1:]
k=k+1
print(k)
``` | output | 1 | 91,247 | 8 | 182,495 |
Provide tags and a correct Python 3 solution for this coding contest problem.
Fox Ciel has n boxes in her room. They have the same size and weight, but they might have different strength. The i-th box can hold at most xi boxes on its top (we'll call xi the strength of the box).
Since all the boxes have the same size, Ciel cannot put more than one box directly on the top of some box. For example, imagine Ciel has three boxes: the first has strength 2, the second has strength 1 and the third has strength 1. She cannot put the second and the third box simultaneously directly on the top of the first one. But she can put the second box directly on the top of the first one, and then the third box directly on the top of the second one. We will call such a construction of boxes a pile.
<image>
Fox Ciel wants to construct piles from all the boxes. Each pile will contain some boxes from top to bottom, and there cannot be more than xi boxes on the top of i-th box. What is the minimal number of piles she needs to construct?
Input
The first line contains an integer n (1 ≤ n ≤ 100). The next line contains n integers x1, x2, ..., xn (0 ≤ xi ≤ 100).
Output
Output a single integer — the minimal possible number of piles.
Examples
Input
3
0 0 10
Output
2
Input
5
0 1 2 3 4
Output
1
Input
4
0 0 0 0
Output
4
Input
9
0 1 0 2 0 1 1 2 10
Output
3
Note
In example 1, one optimal way is to build 2 piles: the first pile contains boxes 1 and 3 (from top to bottom), the second pile contains only box 2.
<image>
In example 2, we can build only 1 pile that contains boxes 1, 2, 3, 4, 5 (from top to bottom).
<image> | instruction | 0 | 91,248 | 8 | 182,496 |
Tags: greedy, sortings
Correct Solution:
```
N = int(input())
ar = list(map(int, input().split()))
ar.sort()
a = 1
b = 100
while a < b:
compliant = True
k = (a+b) // 2
for i in range(len(ar)):
if ar[i] < (i//k):
compliant = False
break
if compliant:
b = k
else:
a = k + 1
print(b)
``` | output | 1 | 91,248 | 8 | 182,497 |
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response.
Fox Ciel has n boxes in her room. They have the same size and weight, but they might have different strength. The i-th box can hold at most xi boxes on its top (we'll call xi the strength of the box).
Since all the boxes have the same size, Ciel cannot put more than one box directly on the top of some box. For example, imagine Ciel has three boxes: the first has strength 2, the second has strength 1 and the third has strength 1. She cannot put the second and the third box simultaneously directly on the top of the first one. But she can put the second box directly on the top of the first one, and then the third box directly on the top of the second one. We will call such a construction of boxes a pile.
<image>
Fox Ciel wants to construct piles from all the boxes. Each pile will contain some boxes from top to bottom, and there cannot be more than xi boxes on the top of i-th box. What is the minimal number of piles she needs to construct?
Input
The first line contains an integer n (1 ≤ n ≤ 100). The next line contains n integers x1, x2, ..., xn (0 ≤ xi ≤ 100).
Output
Output a single integer — the minimal possible number of piles.
Examples
Input
3
0 0 10
Output
2
Input
5
0 1 2 3 4
Output
1
Input
4
0 0 0 0
Output
4
Input
9
0 1 0 2 0 1 1 2 10
Output
3
Note
In example 1, one optimal way is to build 2 piles: the first pile contains boxes 1 and 3 (from top to bottom), the second pile contains only box 2.
<image>
In example 2, we can build only 1 pile that contains boxes 1, 2, 3, 4, 5 (from top to bottom).
<image>
Submitted Solution:
```
n = int(input())
arr = list(map(int, input().split()))
arr.sort()
res = 1
for i in range(n):
if arr[i] < i // res:
res += 1
print(res)
``` | instruction | 0 | 91,249 | 8 | 182,498 |
Yes | output | 1 | 91,249 | 8 | 182,499 |
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response.
Fox Ciel has n boxes in her room. They have the same size and weight, but they might have different strength. The i-th box can hold at most xi boxes on its top (we'll call xi the strength of the box).
Since all the boxes have the same size, Ciel cannot put more than one box directly on the top of some box. For example, imagine Ciel has three boxes: the first has strength 2, the second has strength 1 and the third has strength 1. She cannot put the second and the third box simultaneously directly on the top of the first one. But she can put the second box directly on the top of the first one, and then the third box directly on the top of the second one. We will call such a construction of boxes a pile.
<image>
Fox Ciel wants to construct piles from all the boxes. Each pile will contain some boxes from top to bottom, and there cannot be more than xi boxes on the top of i-th box. What is the minimal number of piles she needs to construct?
Input
The first line contains an integer n (1 ≤ n ≤ 100). The next line contains n integers x1, x2, ..., xn (0 ≤ xi ≤ 100).
Output
Output a single integer — the minimal possible number of piles.
Examples
Input
3
0 0 10
Output
2
Input
5
0 1 2 3 4
Output
1
Input
4
0 0 0 0
Output
4
Input
9
0 1 0 2 0 1 1 2 10
Output
3
Note
In example 1, one optimal way is to build 2 piles: the first pile contains boxes 1 and 3 (from top to bottom), the second pile contains only box 2.
<image>
In example 2, we can build only 1 pile that contains boxes 1, 2, 3, 4, 5 (from top to bottom).
<image>
Submitted Solution:
```
n = int(input())
a = list(map(int,input().split()))
a.sort()
res = 0
for i in range(n):
cnt = i+1
lvl = a[i]+1
res = max(res , (cnt+lvl-1)//lvl)
print(res)
``` | instruction | 0 | 91,250 | 8 | 182,500 |
Yes | output | 1 | 91,250 | 8 | 182,501 |
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response.
Fox Ciel has n boxes in her room. They have the same size and weight, but they might have different strength. The i-th box can hold at most xi boxes on its top (we'll call xi the strength of the box).
Since all the boxes have the same size, Ciel cannot put more than one box directly on the top of some box. For example, imagine Ciel has three boxes: the first has strength 2, the second has strength 1 and the third has strength 1. She cannot put the second and the third box simultaneously directly on the top of the first one. But she can put the second box directly on the top of the first one, and then the third box directly on the top of the second one. We will call such a construction of boxes a pile.
<image>
Fox Ciel wants to construct piles from all the boxes. Each pile will contain some boxes from top to bottom, and there cannot be more than xi boxes on the top of i-th box. What is the minimal number of piles she needs to construct?
Input
The first line contains an integer n (1 ≤ n ≤ 100). The next line contains n integers x1, x2, ..., xn (0 ≤ xi ≤ 100).
Output
Output a single integer — the minimal possible number of piles.
Examples
Input
3
0 0 10
Output
2
Input
5
0 1 2 3 4
Output
1
Input
4
0 0 0 0
Output
4
Input
9
0 1 0 2 0 1 1 2 10
Output
3
Note
In example 1, one optimal way is to build 2 piles: the first pile contains boxes 1 and 3 (from top to bottom), the second pile contains only box 2.
<image>
In example 2, we can build only 1 pile that contains boxes 1, 2, 3, 4, 5 (from top to bottom).
<image>
Submitted Solution:
```
from sys import stdin
inFile = stdin
tokens = []
tokens_next = 0
def next_str():
global tokens, tokens_next
while tokens_next >= len(tokens):
tokens = inFile.readline().split()
tokens_next = 0
tokens_next += 1
return tokens[tokens_next - 1]
def nextInt():
return int(next_str())
def check(a, n):
# a must be sorted in decresing order
if n == 0:
return 0
if len(a) <= n:
return 1
l = [[i] for i in a[:n]]
allowed = [i for i in a[:n]]
ind = 0
for i in a[n:]:
ind += 1
ind %= len(allowed)
starting_pos = ind
while allowed[ind] == 0:
ind += 1
ind %= len(l)
if ind == starting_pos:
# print(l, 0)
return 0
l[ind] += [i]
allowed[ind] = min(i, allowed[ind] - 1)
# print(l, 1)
return 1
def solve(a):
a.sort(reverse=1)
n = len(a)
low = 0
high = n
while low + 1 < high:
m = (low + high) // 2
if check(a, m):
high = m
else :
low = m
return high
n = nextInt()
a = [nextInt() for i in range(n)]
print(solve(a))
``` | instruction | 0 | 91,251 | 8 | 182,502 |
Yes | output | 1 | 91,251 | 8 | 182,503 |
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response.
Fox Ciel has n boxes in her room. They have the same size and weight, but they might have different strength. The i-th box can hold at most xi boxes on its top (we'll call xi the strength of the box).
Since all the boxes have the same size, Ciel cannot put more than one box directly on the top of some box. For example, imagine Ciel has three boxes: the first has strength 2, the second has strength 1 and the third has strength 1. She cannot put the second and the third box simultaneously directly on the top of the first one. But she can put the second box directly on the top of the first one, and then the third box directly on the top of the second one. We will call such a construction of boxes a pile.
<image>
Fox Ciel wants to construct piles from all the boxes. Each pile will contain some boxes from top to bottom, and there cannot be more than xi boxes on the top of i-th box. What is the minimal number of piles she needs to construct?
Input
The first line contains an integer n (1 ≤ n ≤ 100). The next line contains n integers x1, x2, ..., xn (0 ≤ xi ≤ 100).
Output
Output a single integer — the minimal possible number of piles.
Examples
Input
3
0 0 10
Output
2
Input
5
0 1 2 3 4
Output
1
Input
4
0 0 0 0
Output
4
Input
9
0 1 0 2 0 1 1 2 10
Output
3
Note
In example 1, one optimal way is to build 2 piles: the first pile contains boxes 1 and 3 (from top to bottom), the second pile contains only box 2.
<image>
In example 2, we can build only 1 pile that contains boxes 1, 2, 3, 4, 5 (from top to bottom).
<image>
Submitted Solution:
```
# coding: utf-8
n = int(input())
ans = 0
li1 = [int(i) for i in input().split()]
li2 = []
while li1:
li1.sort()
n = len(li1)
i = 0
while i < n:
if li1[i] < i:
li2.append(li1[i])
del(li1[i])
n -= 1
else:
i += 1
ans += 1
li1 = li2
li2 = []
print(ans)
``` | instruction | 0 | 91,252 | 8 | 182,504 |
Yes | output | 1 | 91,252 | 8 | 182,505 |
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response.
Fox Ciel has n boxes in her room. They have the same size and weight, but they might have different strength. The i-th box can hold at most xi boxes on its top (we'll call xi the strength of the box).
Since all the boxes have the same size, Ciel cannot put more than one box directly on the top of some box. For example, imagine Ciel has three boxes: the first has strength 2, the second has strength 1 and the third has strength 1. She cannot put the second and the third box simultaneously directly on the top of the first one. But she can put the second box directly on the top of the first one, and then the third box directly on the top of the second one. We will call such a construction of boxes a pile.
<image>
Fox Ciel wants to construct piles from all the boxes. Each pile will contain some boxes from top to bottom, and there cannot be more than xi boxes on the top of i-th box. What is the minimal number of piles she needs to construct?
Input
The first line contains an integer n (1 ≤ n ≤ 100). The next line contains n integers x1, x2, ..., xn (0 ≤ xi ≤ 100).
Output
Output a single integer — the minimal possible number of piles.
Examples
Input
3
0 0 10
Output
2
Input
5
0 1 2 3 4
Output
1
Input
4
0 0 0 0
Output
4
Input
9
0 1 0 2 0 1 1 2 10
Output
3
Note
In example 1, one optimal way is to build 2 piles: the first pile contains boxes 1 and 3 (from top to bottom), the second pile contains only box 2.
<image>
In example 2, we can build only 1 pile that contains boxes 1, 2, 3, 4, 5 (from top to bottom).
<image>
Submitted Solution:
```
import sys
input = sys.stdin.readline
'''
'''
li = lambda: list(map(int, input().split()))
n = int(input())
x = li()
x.sort()
piles = []
while x:
xi = x.pop()
best_pile = -1
best_index = None
for i, pile in enumerate(piles):
if pile > best_pile and pile > 0:
best_pile = pile
best_index = i
if best_index == None:
piles.append(xi)
else:
piles[best_index] = min(xi, best_pile - 1)
print(len(piles))
``` | instruction | 0 | 91,253 | 8 | 182,506 |
No | output | 1 | 91,253 | 8 | 182,507 |
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response.
Fox Ciel has n boxes in her room. They have the same size and weight, but they might have different strength. The i-th box can hold at most xi boxes on its top (we'll call xi the strength of the box).
Since all the boxes have the same size, Ciel cannot put more than one box directly on the top of some box. For example, imagine Ciel has three boxes: the first has strength 2, the second has strength 1 and the third has strength 1. She cannot put the second and the third box simultaneously directly on the top of the first one. But she can put the second box directly on the top of the first one, and then the third box directly on the top of the second one. We will call such a construction of boxes a pile.
<image>
Fox Ciel wants to construct piles from all the boxes. Each pile will contain some boxes from top to bottom, and there cannot be more than xi boxes on the top of i-th box. What is the minimal number of piles she needs to construct?
Input
The first line contains an integer n (1 ≤ n ≤ 100). The next line contains n integers x1, x2, ..., xn (0 ≤ xi ≤ 100).
Output
Output a single integer — the minimal possible number of piles.
Examples
Input
3
0 0 10
Output
2
Input
5
0 1 2 3 4
Output
1
Input
4
0 0 0 0
Output
4
Input
9
0 1 0 2 0 1 1 2 10
Output
3
Note
In example 1, one optimal way is to build 2 piles: the first pile contains boxes 1 and 3 (from top to bottom), the second pile contains only box 2.
<image>
In example 2, we can build only 1 pile that contains boxes 1, 2, 3, 4, 5 (from top to bottom).
<image>
Submitted Solution:
```
from collections import defaultdict as dc
def mlt(): return map(int, input().split())
vis = dc(lambda: 0)
x = int(input())
s = [*mlt()]
res = 0
for n in s:
vis[n] += 1
res = max(res, vis[n])
print(res)
``` | instruction | 0 | 91,254 | 8 | 182,508 |
No | output | 1 | 91,254 | 8 | 182,509 |
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response.
Fox Ciel has n boxes in her room. They have the same size and weight, but they might have different strength. The i-th box can hold at most xi boxes on its top (we'll call xi the strength of the box).
Since all the boxes have the same size, Ciel cannot put more than one box directly on the top of some box. For example, imagine Ciel has three boxes: the first has strength 2, the second has strength 1 and the third has strength 1. She cannot put the second and the third box simultaneously directly on the top of the first one. But she can put the second box directly on the top of the first one, and then the third box directly on the top of the second one. We will call such a construction of boxes a pile.
<image>
Fox Ciel wants to construct piles from all the boxes. Each pile will contain some boxes from top to bottom, and there cannot be more than xi boxes on the top of i-th box. What is the minimal number of piles she needs to construct?
Input
The first line contains an integer n (1 ≤ n ≤ 100). The next line contains n integers x1, x2, ..., xn (0 ≤ xi ≤ 100).
Output
Output a single integer — the minimal possible number of piles.
Examples
Input
3
0 0 10
Output
2
Input
5
0 1 2 3 4
Output
1
Input
4
0 0 0 0
Output
4
Input
9
0 1 0 2 0 1 1 2 10
Output
3
Note
In example 1, one optimal way is to build 2 piles: the first pile contains boxes 1 and 3 (from top to bottom), the second pile contains only box 2.
<image>
In example 2, we can build only 1 pile that contains boxes 1, 2, 3, 4, 5 (from top to bottom).
<image>
Submitted Solution:
```
n=int(input())
arr=[int(i) for i in input().split()]
arr.sort()
ans=0
while(arr!=[]):
ans+=1
c=arr[-1]
del arr[-1]
if len(arr)==0:
break
while(c>0 and arr!=[]):
boo=False
for i in range(len(arr)-1,-1,-1):
if arr[i]<c:
boo=True
c=min(arr[i],c-1)
del arr[i]
break
if boo==False:
c-=1
del arr[-1]
print(ans)
``` | instruction | 0 | 91,255 | 8 | 182,510 |
No | output | 1 | 91,255 | 8 | 182,511 |
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response.
Fox Ciel has n boxes in her room. They have the same size and weight, but they might have different strength. The i-th box can hold at most xi boxes on its top (we'll call xi the strength of the box).
Since all the boxes have the same size, Ciel cannot put more than one box directly on the top of some box. For example, imagine Ciel has three boxes: the first has strength 2, the second has strength 1 and the third has strength 1. She cannot put the second and the third box simultaneously directly on the top of the first one. But she can put the second box directly on the top of the first one, and then the third box directly on the top of the second one. We will call such a construction of boxes a pile.
<image>
Fox Ciel wants to construct piles from all the boxes. Each pile will contain some boxes from top to bottom, and there cannot be more than xi boxes on the top of i-th box. What is the minimal number of piles she needs to construct?
Input
The first line contains an integer n (1 ≤ n ≤ 100). The next line contains n integers x1, x2, ..., xn (0 ≤ xi ≤ 100).
Output
Output a single integer — the minimal possible number of piles.
Examples
Input
3
0 0 10
Output
2
Input
5
0 1 2 3 4
Output
1
Input
4
0 0 0 0
Output
4
Input
9
0 1 0 2 0 1 1 2 10
Output
3
Note
In example 1, one optimal way is to build 2 piles: the first pile contains boxes 1 and 3 (from top to bottom), the second pile contains only box 2.
<image>
In example 2, we can build only 1 pile that contains boxes 1, 2, 3, 4, 5 (from top to bottom).
<image>
Submitted Solution:
```
# Problem: A. Fox and Box Accumulation
# Contest: Codeforces - Codeforces Round #228 (Div. 1)
# URL: https://codeforces.com/problemset/problem/388/A
# Memory Limit: 256 MB
# Time Limit: 1000 ms
#
# KAPOOR'S
from sys import stdin, stdout
def INI():
return int(stdin.readline())
def INL():
return [int(_) for _ in stdin.readline().split()]
def INS():
return stdin.readline()
def MOD():
return pow(10,9)+7
def OPS(ans):
stdout.write(str(ans)+"\n")
def OPL(ans):
[stdout.write(str(_)+" ") for _ in ans]
stdout.write("\n")
import math
if __name__=="__main__":
n=INI()
X=sorted(INL())
D=dict()
for _ in X:
D[_]=D.get(_,0)+1
ans=0
for _ in D:
ans+=math.ceil((D[_]-ans)/(_+1))
OPS(ans)
``` | instruction | 0 | 91,256 | 8 | 182,512 |
No | output | 1 | 91,256 | 8 | 182,513 |
Provide tags and a correct Python 2 solution for this coding contest problem.
One day, little Vasya found himself in a maze consisting of (n + 1) rooms, numbered from 1 to (n + 1). Initially, Vasya is at the first room and to get out of the maze, he needs to get to the (n + 1)-th one.
The maze is organized as follows. Each room of the maze has two one-way portals. Let's consider room number i (1 ≤ i ≤ n), someone can use the first portal to move from it to room number (i + 1), also someone can use the second portal to move from it to room number pi, where 1 ≤ pi ≤ i.
In order not to get lost, Vasya decided to act as follows.
* Each time Vasya enters some room, he paints a cross on its ceiling. Initially, Vasya paints a cross at the ceiling of room 1.
* Let's assume that Vasya is in room i and has already painted a cross on its ceiling. Then, if the ceiling now contains an odd number of crosses, Vasya uses the second portal (it leads to room pi), otherwise Vasya uses the first portal.
Help Vasya determine the number of times he needs to use portals to get to room (n + 1) in the end.
Input
The first line contains integer n (1 ≤ n ≤ 103) — the number of rooms. The second line contains n integers pi (1 ≤ pi ≤ i). Each pi denotes the number of the room, that someone can reach, if he will use the second portal in the i-th room.
Output
Print a single number — the number of portal moves the boy needs to go out of the maze. As the number can be rather large, print it modulo 1000000007 (109 + 7).
Examples
Input
2
1 2
Output
4
Input
4
1 1 2 3
Output
20
Input
5
1 1 1 1 1
Output
62 | instruction | 0 | 91,257 | 8 | 182,514 |
Tags: dp, implementation
Correct Solution:
```
from sys import stdin, stdout
from collections import Counter, defaultdict
from itertools import permutations, combinations
raw_input = stdin.readline
pr = stdout.write
def in_arr():
return map(int,raw_input().split())
def pr_num(n):
stdout.write(str(n)+'\n')
def pr_arr(arr):
pr(' '.join(map(str,arr))+'\n')
range = xrange # not for python 3.0+
# main code
mod=1000000007
n=int(raw_input())
l=in_arr()
sm=2
dp=[0]*n
dp[0]=2
for i in range(1,n):
if l[i]==i+1:
dp[i]=2
sm=(sm+2)%mod
continue
temp=0
for j in range(l[i]-1,i):
temp=(temp+dp[j])%mod
dp[i]=(temp+2)%mod
sm=(sm+dp[i])%mod
pr_num(sm)
``` | output | 1 | 91,257 | 8 | 182,515 |
Provide tags and a correct Python 3 solution for this coding contest problem.
One day, little Vasya found himself in a maze consisting of (n + 1) rooms, numbered from 1 to (n + 1). Initially, Vasya is at the first room and to get out of the maze, he needs to get to the (n + 1)-th one.
The maze is organized as follows. Each room of the maze has two one-way portals. Let's consider room number i (1 ≤ i ≤ n), someone can use the first portal to move from it to room number (i + 1), also someone can use the second portal to move from it to room number pi, where 1 ≤ pi ≤ i.
In order not to get lost, Vasya decided to act as follows.
* Each time Vasya enters some room, he paints a cross on its ceiling. Initially, Vasya paints a cross at the ceiling of room 1.
* Let's assume that Vasya is in room i and has already painted a cross on its ceiling. Then, if the ceiling now contains an odd number of crosses, Vasya uses the second portal (it leads to room pi), otherwise Vasya uses the first portal.
Help Vasya determine the number of times he needs to use portals to get to room (n + 1) in the end.
Input
The first line contains integer n (1 ≤ n ≤ 103) — the number of rooms. The second line contains n integers pi (1 ≤ pi ≤ i). Each pi denotes the number of the room, that someone can reach, if he will use the second portal in the i-th room.
Output
Print a single number — the number of portal moves the boy needs to go out of the maze. As the number can be rather large, print it modulo 1000000007 (109 + 7).
Examples
Input
2
1 2
Output
4
Input
4
1 1 2 3
Output
20
Input
5
1 1 1 1 1
Output
62 | instruction | 0 | 91,258 | 8 | 182,516 |
Tags: dp, implementation
Correct Solution:
```
R = lambda: map(int, input().split())
n = int(input())
mod = 1000000007
arr = [x - 1 for x in R()]
dp = [0] * (n + 1)
sdp = [0] * (n + 1)
dp[0] = sdp[0] = 2
for i in range(1, len(arr)):
dp[i] = (2 + sdp[i - 1] - sdp[arr[i] - 1]) % mod
sdp[i] = (sdp[i - 1] + dp[i]) % mod
print(sdp[n - 1])
``` | output | 1 | 91,258 | 8 | 182,517 |
Provide tags and a correct Python 3 solution for this coding contest problem.
One day, little Vasya found himself in a maze consisting of (n + 1) rooms, numbered from 1 to (n + 1). Initially, Vasya is at the first room and to get out of the maze, he needs to get to the (n + 1)-th one.
The maze is organized as follows. Each room of the maze has two one-way portals. Let's consider room number i (1 ≤ i ≤ n), someone can use the first portal to move from it to room number (i + 1), also someone can use the second portal to move from it to room number pi, where 1 ≤ pi ≤ i.
In order not to get lost, Vasya decided to act as follows.
* Each time Vasya enters some room, he paints a cross on its ceiling. Initially, Vasya paints a cross at the ceiling of room 1.
* Let's assume that Vasya is in room i and has already painted a cross on its ceiling. Then, if the ceiling now contains an odd number of crosses, Vasya uses the second portal (it leads to room pi), otherwise Vasya uses the first portal.
Help Vasya determine the number of times he needs to use portals to get to room (n + 1) in the end.
Input
The first line contains integer n (1 ≤ n ≤ 103) — the number of rooms. The second line contains n integers pi (1 ≤ pi ≤ i). Each pi denotes the number of the room, that someone can reach, if he will use the second portal in the i-th room.
Output
Print a single number — the number of portal moves the boy needs to go out of the maze. As the number can be rather large, print it modulo 1000000007 (109 + 7).
Examples
Input
2
1 2
Output
4
Input
4
1 1 2 3
Output
20
Input
5
1 1 1 1 1
Output
62 | instruction | 0 | 91,259 | 8 | 182,518 |
Tags: dp, implementation
Correct Solution:
```
n = int(input())
a = list(map(int, input().split()))
dp = [0 for i in range(n+1)]
for i in range(n+1):
if i > 0 : dp[i] = (2*dp[i-1]+2-dp[a[i-1]-1])%1000000007
print((dp[n]+1000000007)%1000000007)
``` | output | 1 | 91,259 | 8 | 182,519 |
Provide tags and a correct Python 3 solution for this coding contest problem.
One day, little Vasya found himself in a maze consisting of (n + 1) rooms, numbered from 1 to (n + 1). Initially, Vasya is at the first room and to get out of the maze, he needs to get to the (n + 1)-th one.
The maze is organized as follows. Each room of the maze has two one-way portals. Let's consider room number i (1 ≤ i ≤ n), someone can use the first portal to move from it to room number (i + 1), also someone can use the second portal to move from it to room number pi, where 1 ≤ pi ≤ i.
In order not to get lost, Vasya decided to act as follows.
* Each time Vasya enters some room, he paints a cross on its ceiling. Initially, Vasya paints a cross at the ceiling of room 1.
* Let's assume that Vasya is in room i and has already painted a cross on its ceiling. Then, if the ceiling now contains an odd number of crosses, Vasya uses the second portal (it leads to room pi), otherwise Vasya uses the first portal.
Help Vasya determine the number of times he needs to use portals to get to room (n + 1) in the end.
Input
The first line contains integer n (1 ≤ n ≤ 103) — the number of rooms. The second line contains n integers pi (1 ≤ pi ≤ i). Each pi denotes the number of the room, that someone can reach, if he will use the second portal in the i-th room.
Output
Print a single number — the number of portal moves the boy needs to go out of the maze. As the number can be rather large, print it modulo 1000000007 (109 + 7).
Examples
Input
2
1 2
Output
4
Input
4
1 1 2 3
Output
20
Input
5
1 1 1 1 1
Output
62 | instruction | 0 | 91,260 | 8 | 182,520 |
Tags: dp, implementation
Correct Solution:
```
import sys
MOD = 10 ** 9 + 7
N = int(input())
bs = [int(b) - 1 for b in input().split()]
fs = [0]
for i in range(1,N+1):
f = 2 * fs[i - 1] + 2 - fs[bs[i - 1]]
fs.append(f % MOD)
print(fs[-1])
``` | output | 1 | 91,260 | 8 | 182,521 |
Provide tags and a correct Python 3 solution for this coding contest problem.
One day, little Vasya found himself in a maze consisting of (n + 1) rooms, numbered from 1 to (n + 1). Initially, Vasya is at the first room and to get out of the maze, he needs to get to the (n + 1)-th one.
The maze is organized as follows. Each room of the maze has two one-way portals. Let's consider room number i (1 ≤ i ≤ n), someone can use the first portal to move from it to room number (i + 1), also someone can use the second portal to move from it to room number pi, where 1 ≤ pi ≤ i.
In order not to get lost, Vasya decided to act as follows.
* Each time Vasya enters some room, he paints a cross on its ceiling. Initially, Vasya paints a cross at the ceiling of room 1.
* Let's assume that Vasya is in room i and has already painted a cross on its ceiling. Then, if the ceiling now contains an odd number of crosses, Vasya uses the second portal (it leads to room pi), otherwise Vasya uses the first portal.
Help Vasya determine the number of times he needs to use portals to get to room (n + 1) in the end.
Input
The first line contains integer n (1 ≤ n ≤ 103) — the number of rooms. The second line contains n integers pi (1 ≤ pi ≤ i). Each pi denotes the number of the room, that someone can reach, if he will use the second portal in the i-th room.
Output
Print a single number — the number of portal moves the boy needs to go out of the maze. As the number can be rather large, print it modulo 1000000007 (109 + 7).
Examples
Input
2
1 2
Output
4
Input
4
1 1 2 3
Output
20
Input
5
1 1 1 1 1
Output
62 | instruction | 0 | 91,261 | 8 | 182,522 |
Tags: dp, implementation
Correct Solution:
```
z = int(input())
l = [None] + [int(x) for x in input().split()]
pt = [0] * (z + 2)
for i in range(1, z + 1):
pt[i + 1] = (2 * pt[i] - pt[l[i]] + 2) % 1000000007
print(pt[z + 1] % 1000000007)
``` | output | 1 | 91,261 | 8 | 182,523 |
Provide tags and a correct Python 3 solution for this coding contest problem.
One day, little Vasya found himself in a maze consisting of (n + 1) rooms, numbered from 1 to (n + 1). Initially, Vasya is at the first room and to get out of the maze, he needs to get to the (n + 1)-th one.
The maze is organized as follows. Each room of the maze has two one-way portals. Let's consider room number i (1 ≤ i ≤ n), someone can use the first portal to move from it to room number (i + 1), also someone can use the second portal to move from it to room number pi, where 1 ≤ pi ≤ i.
In order not to get lost, Vasya decided to act as follows.
* Each time Vasya enters some room, he paints a cross on its ceiling. Initially, Vasya paints a cross at the ceiling of room 1.
* Let's assume that Vasya is in room i and has already painted a cross on its ceiling. Then, if the ceiling now contains an odd number of crosses, Vasya uses the second portal (it leads to room pi), otherwise Vasya uses the first portal.
Help Vasya determine the number of times he needs to use portals to get to room (n + 1) in the end.
Input
The first line contains integer n (1 ≤ n ≤ 103) — the number of rooms. The second line contains n integers pi (1 ≤ pi ≤ i). Each pi denotes the number of the room, that someone can reach, if he will use the second portal in the i-th room.
Output
Print a single number — the number of portal moves the boy needs to go out of the maze. As the number can be rather large, print it modulo 1000000007 (109 + 7).
Examples
Input
2
1 2
Output
4
Input
4
1 1 2 3
Output
20
Input
5
1 1 1 1 1
Output
62 | instruction | 0 | 91,262 | 8 | 182,524 |
Tags: dp, implementation
Correct Solution:
```
n=int(input())
p=list(map(int,input().split(" ",n)[:n]))
a=[i+1 for i in range(n)]
dp1=[0]*(n+1)
dp1[1]=2
dp2=[0]*(n+1)
dp2[1]=2
mod=10**9 + 7
for i in range(2,n+1):
k=p[i-1]
an=2
for j in range(k,i):
an+=dp2[j]%mod
dp2[i]=an%mod
dp1[i]=dp2[i]+dp1[i-1]
dp1[i]%=mod
print(dp1[n])
``` | output | 1 | 91,262 | 8 | 182,525 |
Provide tags and a correct Python 3 solution for this coding contest problem.
One day, little Vasya found himself in a maze consisting of (n + 1) rooms, numbered from 1 to (n + 1). Initially, Vasya is at the first room and to get out of the maze, he needs to get to the (n + 1)-th one.
The maze is organized as follows. Each room of the maze has two one-way portals. Let's consider room number i (1 ≤ i ≤ n), someone can use the first portal to move from it to room number (i + 1), also someone can use the second portal to move from it to room number pi, where 1 ≤ pi ≤ i.
In order not to get lost, Vasya decided to act as follows.
* Each time Vasya enters some room, he paints a cross on its ceiling. Initially, Vasya paints a cross at the ceiling of room 1.
* Let's assume that Vasya is in room i and has already painted a cross on its ceiling. Then, if the ceiling now contains an odd number of crosses, Vasya uses the second portal (it leads to room pi), otherwise Vasya uses the first portal.
Help Vasya determine the number of times he needs to use portals to get to room (n + 1) in the end.
Input
The first line contains integer n (1 ≤ n ≤ 103) — the number of rooms. The second line contains n integers pi (1 ≤ pi ≤ i). Each pi denotes the number of the room, that someone can reach, if he will use the second portal in the i-th room.
Output
Print a single number — the number of portal moves the boy needs to go out of the maze. As the number can be rather large, print it modulo 1000000007 (109 + 7).
Examples
Input
2
1 2
Output
4
Input
4
1 1 2 3
Output
20
Input
5
1 1 1 1 1
Output
62 | instruction | 0 | 91,263 | 8 | 182,526 |
Tags: dp, implementation
Correct Solution:
```
input()
dp = [0]
[dp.append((2 * dp[-1] + 2 - dp[u - 1]) % 1000000007) for u in map(int, input().split())]
print (dp[-1])
``` | output | 1 | 91,263 | 8 | 182,527 |
Provide tags and a correct Python 3 solution for this coding contest problem.
One day, little Vasya found himself in a maze consisting of (n + 1) rooms, numbered from 1 to (n + 1). Initially, Vasya is at the first room and to get out of the maze, he needs to get to the (n + 1)-th one.
The maze is organized as follows. Each room of the maze has two one-way portals. Let's consider room number i (1 ≤ i ≤ n), someone can use the first portal to move from it to room number (i + 1), also someone can use the second portal to move from it to room number pi, where 1 ≤ pi ≤ i.
In order not to get lost, Vasya decided to act as follows.
* Each time Vasya enters some room, he paints a cross on its ceiling. Initially, Vasya paints a cross at the ceiling of room 1.
* Let's assume that Vasya is in room i and has already painted a cross on its ceiling. Then, if the ceiling now contains an odd number of crosses, Vasya uses the second portal (it leads to room pi), otherwise Vasya uses the first portal.
Help Vasya determine the number of times he needs to use portals to get to room (n + 1) in the end.
Input
The first line contains integer n (1 ≤ n ≤ 103) — the number of rooms. The second line contains n integers pi (1 ≤ pi ≤ i). Each pi denotes the number of the room, that someone can reach, if he will use the second portal in the i-th room.
Output
Print a single number — the number of portal moves the boy needs to go out of the maze. As the number can be rather large, print it modulo 1000000007 (109 + 7).
Examples
Input
2
1 2
Output
4
Input
4
1 1 2 3
Output
20
Input
5
1 1 1 1 1
Output
62 | instruction | 0 | 91,264 | 8 | 182,528 |
Tags: dp, implementation
Correct Solution:
```
n,a= int(input()),list(map(int,input().split()))
f,m= [0]*(n+1),10**9+7
for i in range(n):
if a[i]==i+1:
f[i+1]=f[i]+2
else:
f[i+1]=(2+f[i]*2-f[a[i]-1])%m
print(f[n]%m)
``` | output | 1 | 91,264 | 8 | 182,529 |
Provide tags and a correct Python 3 solution for this coding contest problem.
One day, little Vasya found himself in a maze consisting of (n + 1) rooms, numbered from 1 to (n + 1). Initially, Vasya is at the first room and to get out of the maze, he needs to get to the (n + 1)-th one.
The maze is organized as follows. Each room of the maze has two one-way portals. Let's consider room number i (1 ≤ i ≤ n), someone can use the first portal to move from it to room number (i + 1), also someone can use the second portal to move from it to room number pi, where 1 ≤ pi ≤ i.
In order not to get lost, Vasya decided to act as follows.
* Each time Vasya enters some room, he paints a cross on its ceiling. Initially, Vasya paints a cross at the ceiling of room 1.
* Let's assume that Vasya is in room i and has already painted a cross on its ceiling. Then, if the ceiling now contains an odd number of crosses, Vasya uses the second portal (it leads to room pi), otherwise Vasya uses the first portal.
Help Vasya determine the number of times he needs to use portals to get to room (n + 1) in the end.
Input
The first line contains integer n (1 ≤ n ≤ 103) — the number of rooms. The second line contains n integers pi (1 ≤ pi ≤ i). Each pi denotes the number of the room, that someone can reach, if he will use the second portal in the i-th room.
Output
Print a single number — the number of portal moves the boy needs to go out of the maze. As the number can be rather large, print it modulo 1000000007 (109 + 7).
Examples
Input
2
1 2
Output
4
Input
4
1 1 2 3
Output
20
Input
5
1 1 1 1 1
Output
62 | instruction | 0 | 91,265 | 8 | 182,530 |
Tags: dp, implementation
Correct Solution:
```
def bf(n, portal2):
portal2.insert(0, -1)
roomMark = [False]*(n+1)
markCount = 0
room = 1
while(room != n+1):
markCount += 1
roomMark[room] = not roomMark[room]
if roomMark[room]:
room = portal2[room]
else:
room += 1
print(markCount % 1000000007)
def solve(n, portal2):
portal2.insert(0, -1)
f = [0, 2]
for i in range(2, n+1):
total = 2
for j in range(portal2[i], i):
total += f[j]
f.append(total)
print(sum(f) % 1000000007)
n = int(input())
portal2 = [int(i) for i in input().split(" ")]
solve(n, portal2.copy())
# bf(n, portal2.copy())
``` | output | 1 | 91,265 | 8 | 182,531 |
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response.
One day, little Vasya found himself in a maze consisting of (n + 1) rooms, numbered from 1 to (n + 1). Initially, Vasya is at the first room and to get out of the maze, he needs to get to the (n + 1)-th one.
The maze is organized as follows. Each room of the maze has two one-way portals. Let's consider room number i (1 ≤ i ≤ n), someone can use the first portal to move from it to room number (i + 1), also someone can use the second portal to move from it to room number pi, where 1 ≤ pi ≤ i.
In order not to get lost, Vasya decided to act as follows.
* Each time Vasya enters some room, he paints a cross on its ceiling. Initially, Vasya paints a cross at the ceiling of room 1.
* Let's assume that Vasya is in room i and has already painted a cross on its ceiling. Then, if the ceiling now contains an odd number of crosses, Vasya uses the second portal (it leads to room pi), otherwise Vasya uses the first portal.
Help Vasya determine the number of times he needs to use portals to get to room (n + 1) in the end.
Input
The first line contains integer n (1 ≤ n ≤ 103) — the number of rooms. The second line contains n integers pi (1 ≤ pi ≤ i). Each pi denotes the number of the room, that someone can reach, if he will use the second portal in the i-th room.
Output
Print a single number — the number of portal moves the boy needs to go out of the maze. As the number can be rather large, print it modulo 1000000007 (109 + 7).
Examples
Input
2
1 2
Output
4
Input
4
1 1 2 3
Output
20
Input
5
1 1 1 1 1
Output
62
Submitted Solution:
```
n = int(input())
p = list(map(int, input().split()))
m = int(1e9 + 7)
dp = [0] * (n + 1)
dp[0] = 0
for i in range(1, n + 1):
dp[i] = (2 * dp[i - 1] - dp[p[i - 1] - 1] + 2) % m
print(dp[n])
``` | instruction | 0 | 91,266 | 8 | 182,532 |
Yes | output | 1 | 91,266 | 8 | 182,533 |
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response.
One day, little Vasya found himself in a maze consisting of (n + 1) rooms, numbered from 1 to (n + 1). Initially, Vasya is at the first room and to get out of the maze, he needs to get to the (n + 1)-th one.
The maze is organized as follows. Each room of the maze has two one-way portals. Let's consider room number i (1 ≤ i ≤ n), someone can use the first portal to move from it to room number (i + 1), also someone can use the second portal to move from it to room number pi, where 1 ≤ pi ≤ i.
In order not to get lost, Vasya decided to act as follows.
* Each time Vasya enters some room, he paints a cross on its ceiling. Initially, Vasya paints a cross at the ceiling of room 1.
* Let's assume that Vasya is in room i and has already painted a cross on its ceiling. Then, if the ceiling now contains an odd number of crosses, Vasya uses the second portal (it leads to room pi), otherwise Vasya uses the first portal.
Help Vasya determine the number of times he needs to use portals to get to room (n + 1) in the end.
Input
The first line contains integer n (1 ≤ n ≤ 103) — the number of rooms. The second line contains n integers pi (1 ≤ pi ≤ i). Each pi denotes the number of the room, that someone can reach, if he will use the second portal in the i-th room.
Output
Print a single number — the number of portal moves the boy needs to go out of the maze. As the number can be rather large, print it modulo 1000000007 (109 + 7).
Examples
Input
2
1 2
Output
4
Input
4
1 1 2 3
Output
20
Input
5
1 1 1 1 1
Output
62
Submitted Solution:
```
n = int(input()) + 1
mas = [int(x) for x in input().split()]
dp = [0] * n
mod = 1000000007
for i in range(1, n):
dp[i] = (sum(dp[mas[i - 1]:i]) + 2) % mod
print(sum(dp) % mod)
``` | instruction | 0 | 91,267 | 8 | 182,534 |
Yes | output | 1 | 91,267 | 8 | 182,535 |
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response.
One day, little Vasya found himself in a maze consisting of (n + 1) rooms, numbered from 1 to (n + 1). Initially, Vasya is at the first room and to get out of the maze, he needs to get to the (n + 1)-th one.
The maze is organized as follows. Each room of the maze has two one-way portals. Let's consider room number i (1 ≤ i ≤ n), someone can use the first portal to move from it to room number (i + 1), also someone can use the second portal to move from it to room number pi, where 1 ≤ pi ≤ i.
In order not to get lost, Vasya decided to act as follows.
* Each time Vasya enters some room, he paints a cross on its ceiling. Initially, Vasya paints a cross at the ceiling of room 1.
* Let's assume that Vasya is in room i and has already painted a cross on its ceiling. Then, if the ceiling now contains an odd number of crosses, Vasya uses the second portal (it leads to room pi), otherwise Vasya uses the first portal.
Help Vasya determine the number of times he needs to use portals to get to room (n + 1) in the end.
Input
The first line contains integer n (1 ≤ n ≤ 103) — the number of rooms. The second line contains n integers pi (1 ≤ pi ≤ i). Each pi denotes the number of the room, that someone can reach, if he will use the second portal in the i-th room.
Output
Print a single number — the number of portal moves the boy needs to go out of the maze. As the number can be rather large, print it modulo 1000000007 (109 + 7).
Examples
Input
2
1 2
Output
4
Input
4
1 1 2 3
Output
20
Input
5
1 1 1 1 1
Output
62
Submitted Solution:
```
mod = 10**9+7
n = int(input())
a = list(map(int,input().split()))
if n==1:
print (2)
exit()
dp = [0 for i in range(n)]
dp[1] = 2
for i in range(2,n):
dp[i] = (2*dp[i-1]+2-dp[a[i-1]-1])%mod
ans = (2*dp[-1]+2-dp[a[-1]-1])%mod
# print (dp)
print (ans)
``` | instruction | 0 | 91,268 | 8 | 182,536 |
Yes | output | 1 | 91,268 | 8 | 182,537 |
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response.
One day, little Vasya found himself in a maze consisting of (n + 1) rooms, numbered from 1 to (n + 1). Initially, Vasya is at the first room and to get out of the maze, he needs to get to the (n + 1)-th one.
The maze is organized as follows. Each room of the maze has two one-way portals. Let's consider room number i (1 ≤ i ≤ n), someone can use the first portal to move from it to room number (i + 1), also someone can use the second portal to move from it to room number pi, where 1 ≤ pi ≤ i.
In order not to get lost, Vasya decided to act as follows.
* Each time Vasya enters some room, he paints a cross on its ceiling. Initially, Vasya paints a cross at the ceiling of room 1.
* Let's assume that Vasya is in room i and has already painted a cross on its ceiling. Then, if the ceiling now contains an odd number of crosses, Vasya uses the second portal (it leads to room pi), otherwise Vasya uses the first portal.
Help Vasya determine the number of times he needs to use portals to get to room (n + 1) in the end.
Input
The first line contains integer n (1 ≤ n ≤ 103) — the number of rooms. The second line contains n integers pi (1 ≤ pi ≤ i). Each pi denotes the number of the room, that someone can reach, if he will use the second portal in the i-th room.
Output
Print a single number — the number of portal moves the boy needs to go out of the maze. As the number can be rather large, print it modulo 1000000007 (109 + 7).
Examples
Input
2
1 2
Output
4
Input
4
1 1 2 3
Output
20
Input
5
1 1 1 1 1
Output
62
Submitted Solution:
```
'''input
4
1 1 2 3
'''
from sys import stdin, stdout
import sys
sys.setrecursionlimit(15000)
# main starts
n = int(stdin.readline().strip())
arr = list(map(int, stdin.readline().split()))
dp = [-1] * (n + 1)
dp[1] = 2
mod = 10 ** 9 + 7
for i in range(2, n + 1):
dp[i] = (sum(dp[arr[i - 1] : i]) + 2) % mod
print(sum(dp[1:]) % (10 ** 9 + 7))
``` | instruction | 0 | 91,269 | 8 | 182,538 |
Yes | output | 1 | 91,269 | 8 | 182,539 |
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response.
One day, little Vasya found himself in a maze consisting of (n + 1) rooms, numbered from 1 to (n + 1). Initially, Vasya is at the first room and to get out of the maze, he needs to get to the (n + 1)-th one.
The maze is organized as follows. Each room of the maze has two one-way portals. Let's consider room number i (1 ≤ i ≤ n), someone can use the first portal to move from it to room number (i + 1), also someone can use the second portal to move from it to room number pi, where 1 ≤ pi ≤ i.
In order not to get lost, Vasya decided to act as follows.
* Each time Vasya enters some room, he paints a cross on its ceiling. Initially, Vasya paints a cross at the ceiling of room 1.
* Let's assume that Vasya is in room i and has already painted a cross on its ceiling. Then, if the ceiling now contains an odd number of crosses, Vasya uses the second portal (it leads to room pi), otherwise Vasya uses the first portal.
Help Vasya determine the number of times he needs to use portals to get to room (n + 1) in the end.
Input
The first line contains integer n (1 ≤ n ≤ 103) — the number of rooms. The second line contains n integers pi (1 ≤ pi ≤ i). Each pi denotes the number of the room, that someone can reach, if he will use the second portal in the i-th room.
Output
Print a single number — the number of portal moves the boy needs to go out of the maze. As the number can be rather large, print it modulo 1000000007 (109 + 7).
Examples
Input
2
1 2
Output
4
Input
4
1 1 2 3
Output
20
Input
5
1 1 1 1 1
Output
62
Submitted Solution:
```
"""
Code of Ayush Tiwari
Codeforces: servermonk
Codechef: ayush572000
"""
import sys
input = sys.stdin.buffer.readline
def solution():
# This is the main code
n=int(input())
l=list(map(int,input().split()))
l=[0]+l
dp=[0]*(n+2)
for i in range(1,n+1):
dp[i+1]=2*dp[i]+2-dp[l[i]]
print(dp[-1])
t=1
for _ in range(t):
solution()
``` | instruction | 0 | 91,270 | 8 | 182,540 |
No | output | 1 | 91,270 | 8 | 182,541 |
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response.
One day, little Vasya found himself in a maze consisting of (n + 1) rooms, numbered from 1 to (n + 1). Initially, Vasya is at the first room and to get out of the maze, he needs to get to the (n + 1)-th one.
The maze is organized as follows. Each room of the maze has two one-way portals. Let's consider room number i (1 ≤ i ≤ n), someone can use the first portal to move from it to room number (i + 1), also someone can use the second portal to move from it to room number pi, where 1 ≤ pi ≤ i.
In order not to get lost, Vasya decided to act as follows.
* Each time Vasya enters some room, he paints a cross on its ceiling. Initially, Vasya paints a cross at the ceiling of room 1.
* Let's assume that Vasya is in room i and has already painted a cross on its ceiling. Then, if the ceiling now contains an odd number of crosses, Vasya uses the second portal (it leads to room pi), otherwise Vasya uses the first portal.
Help Vasya determine the number of times he needs to use portals to get to room (n + 1) in the end.
Input
The first line contains integer n (1 ≤ n ≤ 103) — the number of rooms. The second line contains n integers pi (1 ≤ pi ≤ i). Each pi denotes the number of the room, that someone can reach, if he will use the second portal in the i-th room.
Output
Print a single number — the number of portal moves the boy needs to go out of the maze. As the number can be rather large, print it modulo 1000000007 (109 + 7).
Examples
Input
2
1 2
Output
4
Input
4
1 1 2 3
Output
20
Input
5
1 1 1 1 1
Output
62
Submitted Solution:
```
print("baad me krunga")
``` | instruction | 0 | 91,271 | 8 | 182,542 |
No | output | 1 | 91,271 | 8 | 182,543 |
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response.
One day, little Vasya found himself in a maze consisting of (n + 1) rooms, numbered from 1 to (n + 1). Initially, Vasya is at the first room and to get out of the maze, he needs to get to the (n + 1)-th one.
The maze is organized as follows. Each room of the maze has two one-way portals. Let's consider room number i (1 ≤ i ≤ n), someone can use the first portal to move from it to room number (i + 1), also someone can use the second portal to move from it to room number pi, where 1 ≤ pi ≤ i.
In order not to get lost, Vasya decided to act as follows.
* Each time Vasya enters some room, he paints a cross on its ceiling. Initially, Vasya paints a cross at the ceiling of room 1.
* Let's assume that Vasya is in room i and has already painted a cross on its ceiling. Then, if the ceiling now contains an odd number of crosses, Vasya uses the second portal (it leads to room pi), otherwise Vasya uses the first portal.
Help Vasya determine the number of times he needs to use portals to get to room (n + 1) in the end.
Input
The first line contains integer n (1 ≤ n ≤ 103) — the number of rooms. The second line contains n integers pi (1 ≤ pi ≤ i). Each pi denotes the number of the room, that someone can reach, if he will use the second portal in the i-th room.
Output
Print a single number — the number of portal moves the boy needs to go out of the maze. As the number can be rather large, print it modulo 1000000007 (109 + 7).
Examples
Input
2
1 2
Output
4
Input
4
1 1 2 3
Output
20
Input
5
1 1 1 1 1
Output
62
Submitted Solution:
```
n=int(input())
p=list(map(int,input().split(" ",n)[:n]))
a=[i+1 for i in range(n)]
dp1=[0]*(n+1)
dp1[1]=2
dp2=[0]*(n+1)
dp2[1]=2
for i in range(2,n+1):
k=p[i-1]
an=2
for j in range(k,i):
an+=dp2[j]
dp2[i]=an
dp1[i]=dp2[i]+dp1[i-1]
print(dp1[n])
``` | instruction | 0 | 91,272 | 8 | 182,544 |
No | output | 1 | 91,272 | 8 | 182,545 |
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response.
One day, little Vasya found himself in a maze consisting of (n + 1) rooms, numbered from 1 to (n + 1). Initially, Vasya is at the first room and to get out of the maze, he needs to get to the (n + 1)-th one.
The maze is organized as follows. Each room of the maze has two one-way portals. Let's consider room number i (1 ≤ i ≤ n), someone can use the first portal to move from it to room number (i + 1), also someone can use the second portal to move from it to room number pi, where 1 ≤ pi ≤ i.
In order not to get lost, Vasya decided to act as follows.
* Each time Vasya enters some room, he paints a cross on its ceiling. Initially, Vasya paints a cross at the ceiling of room 1.
* Let's assume that Vasya is in room i and has already painted a cross on its ceiling. Then, if the ceiling now contains an odd number of crosses, Vasya uses the second portal (it leads to room pi), otherwise Vasya uses the first portal.
Help Vasya determine the number of times he needs to use portals to get to room (n + 1) in the end.
Input
The first line contains integer n (1 ≤ n ≤ 103) — the number of rooms. The second line contains n integers pi (1 ≤ pi ≤ i). Each pi denotes the number of the room, that someone can reach, if he will use the second portal in the i-th room.
Output
Print a single number — the number of portal moves the boy needs to go out of the maze. As the number can be rather large, print it modulo 1000000007 (109 + 7).
Examples
Input
2
1 2
Output
4
Input
4
1 1 2 3
Output
20
Input
5
1 1 1 1 1
Output
62
Submitted Solution:
```
n = int(input())
m=list(map(int,input().split()))
c=[0]*n
for i in range(n):
c[i]=2
for j in range(m[i]-1,i):
c[i]+=c[j]
print(sum(c))
``` | instruction | 0 | 91,273 | 8 | 182,546 |
No | output | 1 | 91,273 | 8 | 182,547 |
Evaluate the correctness of the submitted Python 2 solution to the coding contest problem. Provide a "Yes" or "No" response.
One day, little Vasya found himself in a maze consisting of (n + 1) rooms, numbered from 1 to (n + 1). Initially, Vasya is at the first room and to get out of the maze, he needs to get to the (n + 1)-th one.
The maze is organized as follows. Each room of the maze has two one-way portals. Let's consider room number i (1 ≤ i ≤ n), someone can use the first portal to move from it to room number (i + 1), also someone can use the second portal to move from it to room number pi, where 1 ≤ pi ≤ i.
In order not to get lost, Vasya decided to act as follows.
* Each time Vasya enters some room, he paints a cross on its ceiling. Initially, Vasya paints a cross at the ceiling of room 1.
* Let's assume that Vasya is in room i and has already painted a cross on its ceiling. Then, if the ceiling now contains an odd number of crosses, Vasya uses the second portal (it leads to room pi), otherwise Vasya uses the first portal.
Help Vasya determine the number of times he needs to use portals to get to room (n + 1) in the end.
Input
The first line contains integer n (1 ≤ n ≤ 103) — the number of rooms. The second line contains n integers pi (1 ≤ pi ≤ i). Each pi denotes the number of the room, that someone can reach, if he will use the second portal in the i-th room.
Output
Print a single number — the number of portal moves the boy needs to go out of the maze. As the number can be rather large, print it modulo 1000000007 (109 + 7).
Examples
Input
2
1 2
Output
4
Input
4
1 1 2 3
Output
20
Input
5
1 1 1 1 1
Output
62
Submitted Solution:
```
from sys import stdin, stdout
from collections import Counter, defaultdict
from itertools import permutations, combinations
raw_input = stdin.readline
pr = stdout.write
def in_arr():
return map(int,raw_input().split())
def pr_num(n):
stdout.write(str(n)+'\n')
def pr_arr(arr):
pr(' '.join(map(str,arr))+'\n')
range = xrange # not for python 3.0+
# main code
n=int(raw_input())
l=in_arr()
sm=2
dp=[0]*n
dp[0]=2
for i in range(1,n):
if l[i]==i+1:
dp[i]=2
sm+=2
continue
temp=0
for j in range(l[i]-1,i):
temp+=dp[j]
dp[i]=temp+2
sm+=dp[i]
pr_num(sm)
``` | instruction | 0 | 91,274 | 8 | 182,548 |
No | output | 1 | 91,274 | 8 | 182,549 |
Provide a correct Python 3 solution for this coding contest problem.
Takahashi wants to gain muscle, and decides to work out at AtCoder Gym.
The exercise machine at the gym has N buttons, and exactly one of the buttons is lighten up. These buttons are numbered 1 through N. When Button i is lighten up and you press it, the light is turned off, and then Button a_i will be lighten up. It is possible that i=a_i. When Button i is not lighten up, nothing will happen by pressing it.
Initially, Button 1 is lighten up. Takahashi wants to quit pressing buttons when Button 2 is lighten up.
Determine whether this is possible. If the answer is positive, find the minimum number of times he needs to press buttons.
Constraints
* 2 ≤ N ≤ 10^5
* 1 ≤ a_i ≤ N
Input
Input is given from Standard Input in the following format:
N
a_1
a_2
:
a_N
Output
Print -1 if it is impossible to lighten up Button 2. Otherwise, print the minimum number of times we need to press buttons in order to lighten up Button 2.
Examples
Input
3
3
1
2
Output
2
Input
4
3
4
1
2
Output
-1
Input
5
3
3
4
2
4
Output
3 | instruction | 0 | 91,563 | 8 | 183,126 |
"Correct Solution:
```
n = int(input())
a = [int(input()) for i in range(n)]
x = 1
i = 0
while i < n and x != 2:
x = a[x-1]
i += 1
if x == 2:
print(i)
else:
print(-1)
``` | output | 1 | 91,563 | 8 | 183,127 |
Provide a correct Python 3 solution for this coding contest problem.
Takahashi wants to gain muscle, and decides to work out at AtCoder Gym.
The exercise machine at the gym has N buttons, and exactly one of the buttons is lighten up. These buttons are numbered 1 through N. When Button i is lighten up and you press it, the light is turned off, and then Button a_i will be lighten up. It is possible that i=a_i. When Button i is not lighten up, nothing will happen by pressing it.
Initially, Button 1 is lighten up. Takahashi wants to quit pressing buttons when Button 2 is lighten up.
Determine whether this is possible. If the answer is positive, find the minimum number of times he needs to press buttons.
Constraints
* 2 ≤ N ≤ 10^5
* 1 ≤ a_i ≤ N
Input
Input is given from Standard Input in the following format:
N
a_1
a_2
:
a_N
Output
Print -1 if it is impossible to lighten up Button 2. Otherwise, print the minimum number of times we need to press buttons in order to lighten up Button 2.
Examples
Input
3
3
1
2
Output
2
Input
4
3
4
1
2
Output
-1
Input
5
3
3
4
2
4
Output
3 | instruction | 0 | 91,564 | 8 | 183,128 |
"Correct Solution:
```
N=int(input())
a=[int(input()) for i in range(N)]
x=1
for i in range(N):
if(x==2):
print(i)
exit()
x=a[x-1]
print("-1")
``` | output | 1 | 91,564 | 8 | 183,129 |
Provide a correct Python 3 solution for this coding contest problem.
Takahashi wants to gain muscle, and decides to work out at AtCoder Gym.
The exercise machine at the gym has N buttons, and exactly one of the buttons is lighten up. These buttons are numbered 1 through N. When Button i is lighten up and you press it, the light is turned off, and then Button a_i will be lighten up. It is possible that i=a_i. When Button i is not lighten up, nothing will happen by pressing it.
Initially, Button 1 is lighten up. Takahashi wants to quit pressing buttons when Button 2 is lighten up.
Determine whether this is possible. If the answer is positive, find the minimum number of times he needs to press buttons.
Constraints
* 2 ≤ N ≤ 10^5
* 1 ≤ a_i ≤ N
Input
Input is given from Standard Input in the following format:
N
a_1
a_2
:
a_N
Output
Print -1 if it is impossible to lighten up Button 2. Otherwise, print the minimum number of times we need to press buttons in order to lighten up Button 2.
Examples
Input
3
3
1
2
Output
2
Input
4
3
4
1
2
Output
-1
Input
5
3
3
4
2
4
Output
3 | instruction | 0 | 91,565 | 8 | 183,130 |
"Correct Solution:
```
n=int(input())
a=[0]*(n+1)
for i in range(n):
a[i]=int(input())
k=0
ans=-1
for i in range(n):
# print(k)
k=a[k]-1
if(k==1):
ans=i+1
break
print(ans)
``` | output | 1 | 91,565 | 8 | 183,131 |
Provide a correct Python 3 solution for this coding contest problem.
Takahashi wants to gain muscle, and decides to work out at AtCoder Gym.
The exercise machine at the gym has N buttons, and exactly one of the buttons is lighten up. These buttons are numbered 1 through N. When Button i is lighten up and you press it, the light is turned off, and then Button a_i will be lighten up. It is possible that i=a_i. When Button i is not lighten up, nothing will happen by pressing it.
Initially, Button 1 is lighten up. Takahashi wants to quit pressing buttons when Button 2 is lighten up.
Determine whether this is possible. If the answer is positive, find the minimum number of times he needs to press buttons.
Constraints
* 2 ≤ N ≤ 10^5
* 1 ≤ a_i ≤ N
Input
Input is given from Standard Input in the following format:
N
a_1
a_2
:
a_N
Output
Print -1 if it is impossible to lighten up Button 2. Otherwise, print the minimum number of times we need to press buttons in order to lighten up Button 2.
Examples
Input
3
3
1
2
Output
2
Input
4
3
4
1
2
Output
-1
Input
5
3
3
4
2
4
Output
3 | instruction | 0 | 91,566 | 8 | 183,132 |
"Correct Solution:
```
n=int(input())
a=[int(input()) for i in range(n)]
count=1
for i in range(n):
if count == 2:
print(i)
exit()
count=a[count-1]
print(-1)
``` | output | 1 | 91,566 | 8 | 183,133 |
Provide a correct Python 3 solution for this coding contest problem.
Takahashi wants to gain muscle, and decides to work out at AtCoder Gym.
The exercise machine at the gym has N buttons, and exactly one of the buttons is lighten up. These buttons are numbered 1 through N. When Button i is lighten up and you press it, the light is turned off, and then Button a_i will be lighten up. It is possible that i=a_i. When Button i is not lighten up, nothing will happen by pressing it.
Initially, Button 1 is lighten up. Takahashi wants to quit pressing buttons when Button 2 is lighten up.
Determine whether this is possible. If the answer is positive, find the minimum number of times he needs to press buttons.
Constraints
* 2 ≤ N ≤ 10^5
* 1 ≤ a_i ≤ N
Input
Input is given from Standard Input in the following format:
N
a_1
a_2
:
a_N
Output
Print -1 if it is impossible to lighten up Button 2. Otherwise, print the minimum number of times we need to press buttons in order to lighten up Button 2.
Examples
Input
3
3
1
2
Output
2
Input
4
3
4
1
2
Output
-1
Input
5
3
3
4
2
4
Output
3 | instruction | 0 | 91,567 | 8 | 183,134 |
"Correct Solution:
```
n=int(input())
a=[int(input())-1 for _ in range(n)]
i=0
ct=0
while i!=1:
if a[i]==-1:
print("-1")
break
k=i
i=a[i]
a[k]=-1
ct+=1
else:
print(ct)
``` | output | 1 | 91,567 | 8 | 183,135 |
Provide a correct Python 3 solution for this coding contest problem.
Takahashi wants to gain muscle, and decides to work out at AtCoder Gym.
The exercise machine at the gym has N buttons, and exactly one of the buttons is lighten up. These buttons are numbered 1 through N. When Button i is lighten up and you press it, the light is turned off, and then Button a_i will be lighten up. It is possible that i=a_i. When Button i is not lighten up, nothing will happen by pressing it.
Initially, Button 1 is lighten up. Takahashi wants to quit pressing buttons when Button 2 is lighten up.
Determine whether this is possible. If the answer is positive, find the minimum number of times he needs to press buttons.
Constraints
* 2 ≤ N ≤ 10^5
* 1 ≤ a_i ≤ N
Input
Input is given from Standard Input in the following format:
N
a_1
a_2
:
a_N
Output
Print -1 if it is impossible to lighten up Button 2. Otherwise, print the minimum number of times we need to press buttons in order to lighten up Button 2.
Examples
Input
3
3
1
2
Output
2
Input
4
3
4
1
2
Output
-1
Input
5
3
3
4
2
4
Output
3 | instruction | 0 | 91,568 | 8 | 183,136 |
"Correct Solution:
```
n=int(input())
l=[int(input()) for i in range(n)]
count=0
for i in range(n):
if l[count]==2:
print(i+1)
exit()
else:
count=l[count]-1
print(-1)
``` | output | 1 | 91,568 | 8 | 183,137 |
Provide a correct Python 3 solution for this coding contest problem.
Takahashi wants to gain muscle, and decides to work out at AtCoder Gym.
The exercise machine at the gym has N buttons, and exactly one of the buttons is lighten up. These buttons are numbered 1 through N. When Button i is lighten up and you press it, the light is turned off, and then Button a_i will be lighten up. It is possible that i=a_i. When Button i is not lighten up, nothing will happen by pressing it.
Initially, Button 1 is lighten up. Takahashi wants to quit pressing buttons when Button 2 is lighten up.
Determine whether this is possible. If the answer is positive, find the minimum number of times he needs to press buttons.
Constraints
* 2 ≤ N ≤ 10^5
* 1 ≤ a_i ≤ N
Input
Input is given from Standard Input in the following format:
N
a_1
a_2
:
a_N
Output
Print -1 if it is impossible to lighten up Button 2. Otherwise, print the minimum number of times we need to press buttons in order to lighten up Button 2.
Examples
Input
3
3
1
2
Output
2
Input
4
3
4
1
2
Output
-1
Input
5
3
3
4
2
4
Output
3 | instruction | 0 | 91,569 | 8 | 183,138 |
"Correct Solution:
```
n,ans=int(input()),-1
a,m=[int(input())for i in range(n)],1
for i in range(n):
m=a[m-1]
if m==2:
ans=i+1
break
print(ans)
``` | output | 1 | 91,569 | 8 | 183,139 |
Subsets and Splits
No community queries yet
The top public SQL queries from the community will appear here once available.